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Solution to Me N Mine Mathematics VII Oct 2011 By Sunil

– 2 –

CONTENTSCONTENTSCONTENTSCONTENTSCONTENTS

1. Integers

Worksheets (1 to 7) ..................................................................................................... 5

2. Fractions

Worksheets (8 to 14) ................................................................................................. 14

3. Decimals

Worksheets (15 to 21) ............................................................................................... 27

4. Data Handling

Worksheets (22 to 28) ............................................................................................... 37

5. Simple Equations

Worksheets (29 to 36) ............................................................................................... 52

6. Lines and Angles

Worksheets (37 to 43) ............................................................................................... 67

7. Triangles

Worksheets (44 to 50) ............................................................................................... 78

8. Congruence

Worksheets (51 to 56) ............................................................................................... 89

9. Comparing Quantities

Worksheets (57 to 69) ............................................................................................... 97

– 3 –

10. Rational Numbers

Worksheets (70 to 76) ............................................................................................. 116

11. Symmetry and Practical Geometry

Worksheets (77 to 83) ............................................................................................. 128

12. Perimeter and Area

Worksheets (84 to 91) ............................................................................................. 140

13. Algebraic Expressions

Worksheets (92 to 98) ............................................................................................. 155

14. Exponents and Powers

Worksheets (99 to 105) ........................................................................................... 163

15. Visualising Solid Shapes

Worksheets (106 to 111) ......................................................................................... 172

PRACTICE PAPERS (1 to 5) ................................................................................... 179

5ETNI EG SR

WORKSHEET–1

1. (A) (– 21) + (– 29) = – 21 – 29= – (21 + 29)= – (50) = – 50.

2. (C)Let us take all the options one byone.

(A)175 ÷ (–175) = 175–175

= –175175

= – 1.

(B) (– 16) × 10 = – (16 × 10) = – (160)= – 160.

(C)(–70) ÷ (– 10) = – 70–10

=7010

= 7.

3. (B) Clearly, second term

= First term – 3= 10 – 3 = 7

Also, third term = Second term – 3= 7 – 3 = 4

Similarly, fourth term = Third term – 3= 4 – 3 = 1

and fifth term = Fourth term – 3= 1 – 3 = – 2.

4. (B) (– 3) + 7 – (19) = – 3 + 7 – 19= 7 – 3 – 19= 7 – (3 + 19) = 7 – 22= – 15

15 – 8 + (– 9) = 15 – 8 – 9= 15 – (8 + 9)= 15 – 17 = – 2

Clearly, – 15 is less than – 2so, (– 3) + 7 – (19) is less than 15 – 8 + (– 9)∴ (– 3) + 7 – (19) < 15 – 8 + (– 9).

1Chapter

INTEGERS

5. (C)When two negative integers areadded, we always get a negative integer,e.g.,

(– 7) + (– 13) = – 7 – 13 = – (7 + 13)= – 20= a negative integer.

6. (A) On a number line when we add apositive integer, we always move to theright.

7. (B) Let the additive inverse of – 6 is a ,then

– 6 + a = 0 ∴ a = 6.8. (D) 7 + 3 = 10 ≠ – 10.

9. (A) Let us take option (A).

– 3 × 1 = – (3 × 1) = – (3) = – 3

1 × (– 3) = – (1 × 3) = – (3) = – 3

Hence, – 3 × 1 = – 3 = 1 × (– 3) is correct.

10. (C) Let the blank space be filled by a,then

a × (– 9) = – 72 ⇒ – 9a = – 72

⇒ a = – 72– 9

= 729

⇒ a = 8.

11. (B) If in a fraction, 0 is at the place ofdenominator, then the fraction is notdefined.

∴ a ÷ 0 = 0a

is not defined.

12. (B) a ÷ 48 = – 1 or48a

= – 1

or a = – 1 × 48 or a = – 48.13. (A) (– 41) ÷ [(– 40) + (– 1)]

= – 41 ÷ [– 40 – 1) = – 41– 41

= 1.

6 AM T H E M A T C SI VII–

14. (D) The additive identity of everyinteger is 0.

15. (B) As the additive identity of everyinteger is zero, the additive identity of– 23 is 0.

16. (C) Let us take option (C).LHS = (– 12) + 2 + 10 = – 12 + (2 + 10)

= – 12 + 12 = 0RHS = 12 + (– 2) + (– 10) = 12 – 2 – 10

= 12 – (2 + 10) = 12 – 12 = 0Clearly, LHS = RHS.

17. (D) – 212 + 99 – 87 = 99 – 87 – 212= 99 – (87 + 212)= 99 – 299= – 200.

18. (D) Let us take option (D).

[(– 16) ÷ 4] ÷ (– 2) =–16

4

÷ (– 2)

= [– 4] ÷ (– 2)

=– 4– 2

= 2.

which is greater than zero.Hence, [(– 16) ÷ 4] ÷ (– 2) < 0 is incorrect.

19. (D) Since the multiplicative identity ofany integer is 1, therefore, the multipli-cative identity of 7 is 1.

20. (C) We know that addition is commuta-tive for integers, so a + b = b + a is truefor any integers a and b.

WORKSHEET–2

1. All integers between – 2 and 2 are – 1, 0and 1.

2. The successor of – 380 = – 380 + 1

= – 379

The predecessor of – 380 = – 380 – 1

= – 381.

3. In this case, the negative integer mustbe less than – 10. Suppose this is – 16.Now,– 16 + Positive integer = – 10

∴ Positive integer = – 10 – (– 16)= – 10 + 16= + 6.

Hence, the required pair is – 16 and 6.4. (i) 400 + (– 31) + (– 71)

= 400 – 31 – 71= 400 – (31 + 71)= 400 – 102 = 298.

(ii) 937 + (– 37) + 100 + (– 200) + 300= 937 – 37 + 100 – 200 + 300= 937 + 100 + 300 – 37 – 200= (937 + 100 + 300) – (37 + 200)= 1337 – 237 = 1100.

5. (i) First integer = – 27Second integer = – 54

Second integer – First integer= – 54 – (– 27)= – 54 + 27 = – 27.

(ii) First integer = 12 Second integer = – 7Second integer – First integer

= – 7 – (12)= – 7 – 12 = – 19.

6. (i) (– 14) × (– 11) × 10Since the number of negative integersin the product is even (here 2), therefore,their product must be positive.∴ (– 14) × (– 11) × 10 = 14 × 11 × 10

= 154 × 10(∵ 14 × 11 = 154)= 1540.

(ii) (– 4) × (– 5) × (– 2) × (– 1)Since the number of negative integersis even (here 4), so their product mustbe positive.

7ETNI EG SR

∴ (– 4) × (– 5) × (– 2) × (– 1)= 4 × 5 × 2 × 1= 4 × 5 × 2 (∵ 2 × 1 = 2)= 4 × 10 (∵ 5 × 2 = 10)= 40.

7. (– 2 – 5) × (– 6) = (– 7) × (– 6)= 7 × 6 = 42 [∵ (– a) × (– b) = a × b]

(– 2) – 5 × (– 6) = – 2 – [5 × (– 6)]= – 2 – [– 5 × 6] [∵ a × (– b) = – a × b]= – 2 – (– 30)= – 2 + 30 [∵ a – (– b) = a + b]= 28

Clearly, 42 > 28Therefore, (– 2 – 5) × (– 6) is greater.

8. (i) (– 20) ÷ (– 10) = – 20–10

= 2010

= 2.–

=–

a ab b

∵

(ii) (– 15) ÷ (– 3) = –15– 3

= 153

= 5.–

=–

a ab b

∵

9. (i) 20 × 12 + 20 × (– 4) = 20 × (12 – 4)LHS = 20 × 12 + 20 × (– 4)

= 20 × 12 – 20 × 4 [∵ a × (– b) = – a × b]

= 20 × (12 – 4)[∵ a × b – a × c = a × (b – c)]

= RHS. Hence proved.

(ii) 14 × 10 + 14 × (– 20) = 14 × (10 – 20)LHS = 14 × 10 + 14 × (– 20)

= 14 × 10 – 14 × 20[∵ a × (– b) = – a × b]

= 14 × (10 – 20)[∵ a × b – a × c = a × (b – c]


10. (i) 40 ÷ – 1 = 40– 1

= – 40.

(ii) – 37 ÷ (– 1) =

– 371

–1= 37.

WORKSHEET –3

1. (i) 35 ÷ (– 5) = 35– 5

= –355

= – 7.

(ii) 0 × (– 2) = 0.

(iii) – 275 + x = 1 ⇒ x = 1 + 275 = 276.

(iv) (– 59) + 1 = – 59 + 1 = – 58.

2. – 8 on the number line = 8 steps towards the left of 0.+ 3 on the number line = 3 steps towards the right of 0.

∴ – 8 + 3 = 8 steps towards the left of0 and then 3 steps towardsthe right= – 5.

3. The sign of the product depends onlyon the number of negative numbers.

(i) There is even number of negativeintegers, so the product must bepositive.

(ii) There is odd number of negativeintegers, so the product must benegative.

4. There are seven days in a week.

Temperature after the 1st day

= 42 °C – 2 °C = 40 °C

Temperature after the 2nd day

= 40 °C – 2 °C = 38 °C


Temperature after the 3rd day= 38 °C – 2 °C = 36 °C

Temperature after the 4th day= 36 °C – 2 °C = 34 °C




Thus, the temperature after the wholeweek is 28 °C.

5. (i) 120 – (– 80) = 120 + 80 [ä– (–a) = a]= 200.

(ii) 0 – (– 50) = 0 + 50 = 50.

6. (i) [124 × (– 2)] × (– 5)= 124 × [(– 2) × (– 5)]

(Associativity)= 124 × [ 2 × 5]

[∵ (– a) × (– b) = a × b]= 124 × 10 = 1240.

(ii) [(– 1) × {217 × (– 20)}] × 5= [{(– 1) × 217} × (– 20)] × 5

(Associativity)= {(– 217) × (– 20)} × 5= (– 217) × {(– 20) × 5)}

(Associativity)= (– 217) × (– 20 × 5)= (– 217) × (– 100)= 217 × 100 = 21700.

[∵ (– a) × (– b) = a × b]7. 18 × (– 16) + 2 × (– 16) = (– 16) × (18 + 2)

Let us take left hand side.LHS = 18 × (– 16) + 2 × (– 16)

= (18 + 2) × (– 16)[∵ a × c + b × c = (a + b) × c]

= (– 16) × (18 + 2)[Commutativity]

which is RHS.

Let us take right hand side.

RHS = (– 16) × (18 + 2)

= (– 16) × 18 + (– 16) × 2

[Distributivity]

= 18 × (– 16) + 2 × (– 16)

[Commutativity]

which is LHS. Hence proved.

8. a ÷ (b + c) ≠ (a ÷ b) + (a ÷ c)

Let us take LHS of this inequality.

LHS = a ÷ (b + c)

Substituting a = 15, b = – 3 and c = 1,we get

LHS = 15 ÷ (– 3 + 1) = 15 ÷ (– 2)

= 15– 2

= –152

On the same way,

RHS = (a ÷ b) + (a ÷ c)

= [15 ÷ (– 3)] + (15 ÷ 1)

= 15– 3

+

151

= – 5 + 15 = 10.

Clearly, LHS ≠ RHS

i.e., a ÷ (b + c) ≠ (a ÷ b) + (a ÷ c).

9. a ÷ b = – 4

or ab

= – 4 or a = – 4 × b

If b = 1, then a = – 4 × 1 = – 4

If b = 2, then a = – 4 × 2 = – 8

If b = 3, then a = – 4 × 3 = – 12

Thus, three pairs of integers (a, b) are

(– 4, 1), (– 8, 2) and (– 12, 3).

10.1

12× (– 9) = –

912

= – 34

.

9ETNI EG SR

WORKSHEET –4

1. (i)– 25

× 25 × (– 1) = 25 × 25 × 1

[∵ – a × b × (– c) = a × b × c]

= 25

× (25 × 1) = 25

× 25

[∵ a × 1 = a]

= 2 ×255

= 2 × 5 = 10.

(ii)32

× (– 4) × (– 1) = 32

× 4 × 1

= 32

× 4 = 3 ×42

= 3 × 2 = 6.2. Let three negative integers be – 2, – 3

and – 4.Their product = (– 2) × (– 3) × (– 4)

= (– 2) × [(– 3) × (– 4)]

= (– 2) × [3 × 4]

[∵ (– a) × (– b) = a × b]

= (– 2) × 12

[∵ (– a) × b = – (a × b)]

= – (24) = – 24.

= Negative integer.

Hence, the product of three negativeinteger and is a negative integer.

3. (i) – 800000 ÷ (– 200)

= – 800000

– 200 =

800000200

–=

–a ab b

∵

= 8000

2 = 4000.

(ii) 343 ÷ (– 49) = 343– 49

=34349

––a ab b

= ∵

= –497

= – 7.

4. (i) – 4 × 16 × 25 × 3= 16 × (– 4) × 25 × 3 (Commutativity)= 16 × [(– 4) × 25 × 3]= 16 × [{(– 4) × 25} × 3]= 16 × [(– 100) × 3]= [16 × (– 100)] × 3

(Associativity)= – 1600 × 3 = – 4800.

(ii) 4 + (– 8) + 6 + (– 2)= [4 + (– 8) + 6] + (– 2)= [{4 + (– 8)} + 6] + (– 2)= [(– 4) + 6] + (– 2)= (– 4) + [6 + (– 2)]

(Associativity)= – 4 + 4 = 0.

5. Let the other number be a.∴ 60 × a = – 180Dividing both sides by 60, we get

a =–180

60 = –

18060

or a = – 3.

6. Let the number be b.

According to the question, 3b

= 14

Multiplying both sides by 3, we getb = 3 × 14 or b = 42

7. (i) 34 × (– 1) = – (34 × 1)[∵ a × (– b) = – (a × b)]

= – 34. [∵ a × 1 = a] (ii) (– 12) × (– 1) = 12 × 1

[∵ (– a) × (– b) = a × b] = 12. [∵ a × 1 = a]

8. (i) – 66 – (– 22) = – 66 + 22 = – 44. (ii) 100 – (– 42 + 39)

= 100 – (– 42) – (+ 39)= 100 + 42 – 39= 142 – 39 = 103.


9. (i) (– 55) ÷ 11 = – 5511

= –5511

= – 5.

(ii) – 77

7 = –

777

= – 11.

10. 1 hour = 60 minutes2 hours = 2 × 60 minutes

= 120 minutes.∵ In 1 minute the elevator covers adepth of 6 metres∴ In 2 hours the elevator will cover adepth of 6 × 120 metresi.e., 720 metres.Thus, the elevator will be 720 metresbelow the initial position.

11. (i) – 88– 8

=888

= 11.

(ii)– 25

5 = –

255

= – 5.

WORKSHEET –5

1. (i) [4 × (– 112)] × 5= [– (4 × 112)] × 5

[∵ a × (– b) = – (a × b)]= [– (448)] × 5 = (– 448) × 5= – (448 × 5) = – 2240.

(ii) 19 + (– 13 + 3) = 19 + (– 10)= 19 – 10 = 9.

2. (i) 25 × 7 × 4 × 3 = 25 × 4 × 7 × 3= (25 × 4) × (7 × 3)= 100 × 21 = 2100.

(ii) (– 15) + 24 + 5 + (– 4)= (– 15) + 5 + 24 + (– 4)= (– 15 + 5) + (24 – 4)= – 10 + 20 = 10.

3. (i) Additive inverse of 15 = – 15.(ii) Additive inverse of – 23 = 23.(iii) Additive inverse of 0 = 0.

4. (i) 20 × [5 × (– 16)] = (20 × 5) × (– 16)LHS = 20 × [5 × (– 16)] = 20 × [– 80]

= – (20 × 80) = – 1600RHS = (20 × 5) × (– 16) = 100 × (– 16)

= – (100 × 16) = – 1600.So, 20 × [5 × (– 16)] = (20 × 5) × (– 16).

(ii) 18 × [100 + (– 5)] = 18 × 100 + 18 × (– 5)Here 18 × [100 + (– 5)]

= 18 × [100 – 5]= 18 × 95 = 1710

And 18 × 100 + 18 × (– 5)= (18 × 100) – (18 × 5)= 1800 – 90 = 1710.

So, 18 × [100 + (– 5)] = 18 × 100 + 18 × (– 5).

5. Let your home be at O. You was at A.Now, you are at B.AO = 8 km, OB = 4 kmYou travelled from A to B via O.

∴ Required distance travelled by you = AO + OB = 8 km + 4 km.

= 12 km.6. Let the position of bird be at A and the

position of fish at B. AB is a verticalstraight line.

11ETNI EG SR

Now, required distance= AB= 6000 m + 1600 m = 7600 m.

7. a ÷ b = – 3 or ab

= – 3

or a = – 3b[Multiplying both sides by b]

If b = 1, then a = – 3(1) = – 3∴ (a, b) = (– 3, 1).

8. 423 × (– 63) – [63 × (– 423)]= 423 × (– 63) – [(– 423) × 63]

(Commutativity)= – (423 × 63) – [– (423 × 63)]= – (423 × 63) + (423 × 63)

[∵ – a – (– a) = – a + a]= 0.

9. Let one of the two integers be 4. Thenaccording to the question,

4 + another integer = – 20

∴ Another integer = – 20 – 4 = – 24

Hence, the required pair is – 24, 4.

10. (i) 80 × [5 × (– 36)] = (80 × 5) × (– 36)

Let us take left hand side (LHS).

LHS = 80 × [5 × (– 36)]

= (80 × 5) × (– 36)

[Associativity for multiplication]

= RHS Hence proved.

(ii) – 4 × 16 × 25 × 3 = {(– 4) × 25} × (16 × 3)

Let us take left hand side (LHS).

LHS = – 4 × 16 × 25 × 3

= – 4 × (16 × 25) × 3

= – 4 × (25 × 16) × 3[Commutativity of multiplication]

= – 4 × 25 × 16 × 3

= {(– 4) × 25} × (16 × 3)


WORKSHEET –6

1. (i) [13 × 19] × (– 3) = 13 × [19 × (– 3)](Associativity of multiplication)

Thus, the blank space is filled with 19.(ii) (– 10) × 9 × (– 10) × 1

= – 10 × 9 × [(– 10) × 1]= – 10 × 9 × (– 10)

[∵ (– a) × 1 = – a]= – 10 × [9 × (– 10)]= – 10 × [– (9 × 10)]= – 10 × (– 90)

Thus, the blank space is filled with – 90.

2. (i) The additive inverse of – 13 = 13.(ii) The additive inverse of 22 = – 22.

3. 30125 × 99 – (– 30125)= 30125 × 99 + 30125

[∵ – (– a) = a]= 30125 × 99 + 30125 × 1

[∵ a = a × 1]= 30125 × (99 + 1)= 30125 × 100 = 3012500.

4. (i) (– 5) + (– 3) + 2= – 5 – 3 + 2 = – (5 + 3) + 2= – 8 + 2 = – 6.

(ii) (– 613) + (– 111) + (– 500)= – 613 – 111 – 500= – (613 + 111 + 500)= – (1224) = – 1224.

5. The difference of – 19 and – 43= – 19 – (– 43) = – 19 + 43 = 24

Now, required value = – 63 + 24= – 39.

6. Ascending order is– 33, – 10, – 7, – 5, – 3, 0, 4, 6, 11, 19.

7. The product of (– 5) × (6) × (– 7) × (– 20)has an odd number of negative integers,so its value must be negative.


∴ (– 5) × (6) × (– 7) × (– 20)= – 5 × 6 × 7 × 20= – (5 × 20) × (6 × 7)= – 100 × 42 = – 4200.

8. – 4, – 3, – 2, – 1 and 0.

9. Let the one negative integer be –10.

Then, – 10 – (Other negative integer)

= 18

∴ Other negative integer

= – 10 – 18 = – 28

Hence the integers are – 10 and – 28.

10. To find balance finally, we add thedeposits and subtract the withdrawls.∴ So, Anita's balance

= ` 3148 + ` 1500 – ` 2100 + ` 2000 – ` 1550

= ` (3148 + 1500 + 2000) – ` (2100 + 1550)

= ` 6648 – ` 3650 = ` 2, 998.

WORKSHEET –7

1. (i) 738 + (– 99) + 100 – (– 400)= 738 – 99 + 100 + 400= (738 + 100 + 400) – 99= 1238 – 99 = 1139.

(ii) 76 × (– 18) + 76 × 18= 76 × (– 18 + 18)= 76 × 0= 0. [∵ a × 0 = 0]

2. (i) (– 100 + 7) – 63 = – 100 + 7 – 63= 7 – (100 + 63)= 7 – 163 = – 156.

(ii) – 666 – (– 222) = – 666 + 222 [∵ – a – (– b) = – a + b]

= – 444.3. (i) (– 4) × (– 5) × (– 2) × (– 1)

Here the number of negative integers iseven.

∴ (– 4) × (– 5) × (– 2) × (– 1)= 4 × 5 × 2 × 1 = 40.

(ii) 2 × (– 5) × (– 7) × 4Here the number of negative integersis even.∴ 2 × (– 5) × (– 7) × 4

= 2 × 5 × 7 × 4= (2 × 5) × (7 × 4)= 10 × 28 = 280.

(iii) (– 4) × (– 11) × 10Here the number of negative integers iseven.∴ (– 4) × (– 11) × 10 = 4 × 11 × 10 = 440.

4. Difference of 0 and –10 = 0 – (– 10) = 10Sum of 0 and –10 = 0 + (– 10) = – 10

Thus, the required pair is (0, – 10).

5. (i) Rise in the temperature= 6 °C – (– 3 °C)= 6 °C + 3 °C= 9°C.

(ii) The temperature at the end of the afternoon = 5°C – 7°C

= – 2°C.

6. (i) 7009 ÷ (– 7009) = 7009

– 7009

= 7009

–7009

= –

–a ab b

∵

= – 1.

(ii) (– 808) × [110 + (– 33)]= – 808 × [110 – 33]= – 808 × 77 = – 62216.

7. The temperature of water will be 20 °Cafter a change of 20 °C – 80° C = – 60 °C∵ Time taken in the change of – 4°C

= 10 minutes∴ Time taken in the change of – 1°C

= 104

minutes

13ETNI EG SR

∴ Time taken in the change of – 60°C

= 104

× 60 minutes = 150 minutes.

8. We know that the product of a positiveinteger with the negative integer isnegative. So, the required number willbe positive. As twice of the requirednumber is 150, the number will be thehalf of 150.

So, the required number = 150

2 = 75.

9. Required time in hours

= Capacity of the tank

Quantity of water reduced per hour

= 2000 litres

4 litres = 500.

10. (i) 336 × (– 2) × (– 5)= 336 × [(– 2) × (– 5)]= 336 × (2 × 5)

[∵ (– a) × (– b) = a × b]= 336 × 10= 3360.

(ii) 114 × 0 × (– 2) = 114 × [0 × (– 2)] = 114 × 0

[∵ 0 × Any integer = 0] = 0.

❏❏


WORKSHEET–8

1. (A) In 45

, numerator < denominator.

∴ 45

is the proper fraction.

2. (B) 4 – 78

=4 × 8 – 7

8=

32 – 78

=258

= 1

38 .

3. (C) 1

22

+2

33

=4 + 1

2+

9 + 23

=52

+113

=5 × 32 × 3

+11× 23 × 2

[∵ LCM of 2 and 3 = 2 × 3 = 6]

=156

+226

= 376

= 1

66

.

4. (A) 3 ×47

= a

⇒ a = 3 ×47

⇒ a =3 × 4

7

⇒ a =127

⇒ a =5

17

.

5. (D) 34

of 16 = 34

× 16

=3 ×16

4=

484

= 12.

6. (B)14

+14

+14

=34

⇒ 3 ×14

=34

.

2Chapter

FRACTIONS

7. (D) 13

×178

=1×173 × 8

= 1724

.

8. (A) 45

of 521

=45

×521

=4 × 55 × 21

=421

.

9. (C) 215

÷1

15

=10 + 1

5÷

5 + 15

=115

÷ 65

=

11565

=11× 55 × 6

= 116

= 5

16

.

10. (B) ∵ Reciprocal of a non-zero whole

number = 1

Whole number

∴ Reciprocal of a = 1a .

11. (A) Reciprocal of 97

= ( )197

=79

.

12. (A)45

÷ 4 =45

×14

= 4 ×15 × 4

= 15

.

13. (B)12

is a reciprocal of ( )112

= 2.

14. (D) ∵ 35

×53

= 1

Therefore, 35

and 53

are reciprocals of

each other.

15CARF IT NO S

15. (C) Total weight =1

22

kg +1

35 kg

=4 1

2+

kg +15 1

5+

kg

= 52

kg + 165

kg

=5 52 5××

kg +16 25 2××

kg

[∵ LCM of 2 and 5 = 2 × 5 = 10]

=2510

kg + 3210

kg

=5710

kg = 7

510

kg.

16. (B) The distance covered by scooter in1 litre of petrol = 40 kmThe scooter will cover the distance in

33

4 litres of petrol

= 40 × 3

34

km

= 40 ×154

km = 40 15

4×

km

= 10 × 15 km = 150 km.

WORKSHEET –9

1. (i) 213

+1

12

= 2 3 13

× + + 1 2 1

2× +

= 6 1

3+

+2 1

2+

= 73

+32

[∵ LCM of 2 and 3 = 2 × 3 = 6]

∴ 213

+1

12

=7 23 2×× +

3 32 3××

= 146

+96

= 236

= 5

36

.

(ii)5

17

+3

214

= 7 5

7+

+28 3

14+

=127

+3114

=12 2 31 1

14× + ×

[∵ LCM of 7 and 14 = 14]

=24 31

14+

=5514

= 13

314

.

2. (i)9 3

–8 4

2 4, 82 2, 42 1, 2 1, 1

∴ LCM of 4 and 8 = 2 × 2 × 2 = 8

∴ 9 3

–8 4

= 9 1 – 3 2

8× ×

= 9 – 6

8 =

38

.

(ii)1 1

–2 4

2 2, 42 1, 2 1, 1

∴ LCM of 2 and 4 = 2 × 2 = 4

∴ 1 1

–2 4

= 1 2 – 1 1

4× ×

= 2 – 1

4 =

14

.

3. (i) 6

17

× 21 =7 6

7+

× 21 =137

× 21

=13 21

7×

= 13 × 3 = 39.

(ii)3

14

×23

×4

28=

4 34+

×23

×4

28

=74

×23

×4

28

=74

×23

×17

=7 2 14 3 7× ×× ×

=2

4 3× =

12 3×

=16

.


4. (i) 19

of 81 =19

× 81 = 819

= 9.

(ii)14

of 1215

=14

×1215

=1 124 15××

= 3

15 =

15

.

(iii)78

of ` 64 = `7

648

× = `

7 648×

= ` (7 × 8) = ` 56.

5. (i)4

15×

1 54 6

+

2 4, 62 2, 33 1, 3 1, 1

∴ LCM of 4 and 6 = 2 × 2 × 3 = 12

∴ 4 1 5

× +15 4 6

=

415

× ( )1 3 5 212

× + ×

= 4

15×

3 1012+

= 4

15×

1312

= 4 ×13

15 ×12=

1315 × 3

= 1345

.

(ii)4 3

2 + 15 10

× 1

12

=10 + 4 10 + 3

+5 10

×

2 + 12

=14 13

+5 10

×

32

=14 × 2 + 13 ×1

10

×32

=28 + 13

10

×32

=4110 ×

32

=41× 310 × 2

=12320

= 3

620

.

6. (i) Reciprocal of 4 =14

.

(ii) Reciprocal of 25

= ( )125

=52

.

7. (i)15

÷12

=15

×21

=1× 25 ×1

= 25

.

(ii) 18 ÷45

= 18 ×54

=18 × 5

4

=9 × 5

2=

452

= 1

222

.

8. Length of each part =5

12m ÷ 2

=5

12m ×

12

=5 ×1

12 × 2m

=5

24m.

WORKSHEET–10

1. (i) – 218

+– 718

=– 2 – 7

18 =

– 918

= –12

.

(ii) 1125

+– 225

=11 – 2

25 =

925

.

2. (i)–74

+1820

=– 7 × 5 + 18 × 1

20

=– 35 + 18

20 =

–1720

.

(ii)– 57

+4

14=

– 5 × 2 + 4 × 114

=–10 + 4

14

= – 614

= – 37

.

3. (i) –78

+

16

+14

17CARF IT NO S

2 4, 6, 82 22 13 1

, 3, 4, 3, 2, 3, 1

1, 1, 1

So, LCM of 4, 6 and 8 = 2 × 2 × 2 × 3 = 24

Now, –78

+

16

+14

=– 7 × 3 + 1× 4 + 1× 6

24

=– 21 + 4 + 6

24 =

– 1124

.

(ii)13

+–34

+58

2 3, 4, 82 32 33 3

, 2, 4, 1, 2, 1, 1

1, 1, 1

So, LCM of 3, 4 and 8 = 2 × 2 × 2 × 3 = 24

Now, 13

+– 34

+58

=1× 8 – 3 × 6 + 5 × 3

24

=8 – 18 + 15

24=

524

.

4. (i) 910

– 45

LCM of 5 and 10 = 10

∴ 9

10–

45

=9 ×1 – 4 × 2

10 =

9 – 810

= 1

10.

(ii) 7–

18–

29

LCM of 18 and 9 = 18

∴ –718

–29

=– 7 – 4

18 =

–1118

.

5. The given fractions are:

– 611

, –122

, 3

111 ,

72

33

or– 611

, –122

, 11 + 3

11 , 66 + 7

33

or– 611

, –122

, 1411 ,

7333

2 11, 22, 33 3 1111 11 1

, 11, 33, 11, 11

, 1, 1

LCM of 11, 22 and 33 = 2 × 3 × 11 = 66

– 611

=– 6 × 611× 6

= – 3666

–122

=–1× 322 × 3

= – 366

1411

=14 × 611× 6

= 8466

7333

=73 × 233 × 2

= 14666

∵ – 36 < – 3 < 84 < 146

∴ – 3666

<– 366

< 8466 <

14666

i.e, – 611

< – 122

< 3

111 <

72

33 .

6. Perimeter of rectangle= 2 × (length + breadth)

= 2 ×1 3

14 + 102 4

m

= 2 ×29 43

+2 4

m = 2 ×

58 + 434

m

= 2 ×101

4m =

1012

m = 1

502

m.

7. (i) 20 ×15

= 20 ×1

5 = 4.


(ii)1112

× 6 = 11× 6

12 =

112

= 1

52

.

8. (i)34

of 76 =34

× 76 =3 × 76

4

= 3 × 19 = 57.

(ii) 45

of 70 =45

× 70 = 4 × 70

5= 4 × 14 = 56.

9. (i)73 ×

11

3=

73

×3 + 1

3=

73

×43

=7 × 43 × 3

=289

=1

39 .

(ii)38

× 89

=3 × 88 × 9

=39

=13

.

WORKSHEET– 11

1. (i)25 ×

1018

= 2

18 ×

105

= 19

× 2 = 29

Here 2 < 9i.e., numerator < denominator

So, 29 is less than 1.

(ii)73

÷ 6

12 =

73

× 126

= 73

× 2 = 143

Here 14 > 3i.e., numerator > denominator

So, 143

is greater than 1.

2. (i) No.

Example:- 47

is a proper fraction

Reciprocal of 47 is 7

4 or 1

34

Clearly, 134

is not a proper fraction.

(ii) No.

Example:- 115

is an improper fraction

Reciprocal of 115

is 5

11.

Clearly, 511

is not an improper

fraction.

3. (i)89

÷ 415

= 89

× 154

= 84

× 159

= 21

× 53

= 2 51 3××

= 103

= 313

.

(ii) 314

÷ 116

= 3 4 1

4× +

÷ 1 6 1

6× +

= 134

÷ 76

= 134

× 67

= 137

× 64

= 137

× 32

= 13 37 2××

= 3914

= 21114

.

4. (i) 47

of 21 = 47

× 21 = 47

× 211

= 41

× 217

= 41

× 3 = 4 3

1×

= 121

= 12.

(ii) 14 ÷ 75

= 141

× 57

= 147

× 51

= 21

× 51

= 2 51 1××

= 101

= 10.

(iii) 1415

÷ 122

25

19CARF IT NO S

= 14 5 1

5× +

÷12 25 2

25× +

= 715

÷30225

= 715

× 25302

= 71

302 ×

255

= 71

302 ×

51

= 71 5

302×

= 355302

= 153302

.

5. 23

of 2 hours = 23

× 2 = 43

hours

∵ 1 hour = 60 minutes= 60 × 60 seconds

(∵ 1 minute = 60 seconds)

∴43 hours =

43 × 60 × 60 seconds

= 4 × 20 × 60 seconds= 4800 seconds.

6. ∵ In 1 hour Akshit reads = 13

part

∴ In 218 hour he will read =

13

× 218

= 13

× 178

= 1724

part.

So, Akshit read 1724 part of the book.

7. (i) Reciprocal of 35

= ( )135

= 53

.

(ii) Reciprocal of 1211

= ( )1

1211

= 1112

.

8. (i)37 of

16

= 37

× 16

= 1

1435 of

23

= 35

× 23

= 25

LCM of 14 and 5 = 14 × 5 = 70

Now, 1

14=

1 514 5××

= 5

70 (∵

7014

= 5)

and 25

= 2 145 14××

= 2870

(∵ 705

= 14)

∵ 28 > 5 ∴ 2870 >

570

So, 35

of 23

is greater.

(ii) 12

of 89

= 12

× 89

= 49

45

of 1011

= 45

× 1011

= 811

LCM of 9 and 11 = 9 × 11 = 99

Now, 49

=49

×1111

=4499

(∵ 999

= 11)

and 811

= 811

× 99

= 7299

(∵ 9911

= 9)

∵ 72 > 44

∴ 7299 >

4499

So, 45

of 1011

is greater.

9. First we have to find LCM of 3, 9 and12

∴ LCM = 2 × 2 × 3 × 3 = 36.

Now, 13

=13

×1212

=1236

(∵ 363 = 12)

49 =

49

×44

=1636

(∵ 369

= 4)

2 3, 9, 122 3, 9, 63 3, 9, 33 1, 3, 1

1, 1, 1


and 5

12=

512

×33

=1536

(∵ 3612

= 3)

∵ 16 > 15 > 12

∴ 1636 >

1536 >

1236

So, 49

, 512

, 13

is the descending order.

10. Length of main strip = 6 cm

Length of smaller strip = 32

cm

Number of strips

= Length of main strip

Length of smaller strip

= ( )632

= 6 × 23

= 2 × 2

= 4 strips.

WORKSHEET–12

1. (i) 1512 × 6

15

× 11

10

=15 2 + 1

2×

×6 × 5 + 1

5×

1×10 + 110

=30 + 1

2×

30 + 15

×10 + 1

10

=312

×315

×1110

= 31 31 112 5 10× ×× ×

= 10571100

= 10571

100.

(ii)23 of

34 of 26 =

23

×34

× 26

=2 3 26

3 4× ××

=2 26

4×

= 524

= 13.

2. ∵ Cost of 14 litre = ` 20

∴ Cost of 1 litre = 2014

`= ` (20 × 4)

= ` 80

∴ Cost of 1

52

litres = ` 80 ×1

52

= ` 80 ×112

= ` 440.

3. (i)35 × =

2740

Let = a

Then35

× a =2740

⇒ a =2740

×53

=27 540 3

×× =

98

⇒ a =1

18

∴ =1

18

.

(ii)45 + =

1210

Let = b

Then45

+ b =1210

⇒ b =1210

–45

=12 1 – 4 2

10× ×

⇒ b =12 – 8

10= 4

10=

25

∴ =25 .

4. Weight of apples =1

32 kg

=3 2 1

2× +

kg

=72

kg

21CARF IT NO S

Weight of oranges =3

64 kg

=6 4 3

4× +

kg

=274

kg

LCM of 2 and 4 = 4Total weight of fruits= weight of apples + weight of oranges

= 72

kg +274

kg

= 14 27

4+

kg = 414

kg = 1

104

kg.

5.1

23

=2 3 1

3× +

= 73

11

6=

1 6 16

× + =

76

113

12=

3 12 1112

× + =

4712

51

6=

1 6 56

× + =

116

∴ LCM of 3, 6 and 12 = 2 × 2 × 3 = 12

Now, 1

23

+1

16

+11

312

– 5

16

= 73

+76

+4712

–116

= 7 4 7 2 47 1 – 11 2

12× + × + × ×

= 28 14 47 – 22

12+ +

= 6712

=7

512

.

6. AB =3

54

cm = 5 4 3

4× +

cm = 234

cm

BC = 1

82

cm = 8 2 1

2× +

cm = 172

cm

CD = 1

48

cm =4 8 1

8× +

cm = 338

cm

DE =3

124

cm =12 4 3

4× +

= 514

cm.

EA = CD = 338

cm

Now, perimeter of figure ABCDE= AB + BC + CD + DE + EA

=23 17 33 51 334 2 8 4 8

+ + + + cm

LCM of 2, 4 and8 = 2 × 2 × 2 = 8

=23 2 17 4 33 1 51 2 33 1

8× + × + × + × + ×

cm

=46 68 33 102 33

8+ + + +

cm

=2828

cm = 2

354

cm i.e., 1

352

cm.

7. 2 dozen = 2 × 12 = 24

13 of the oranges =

13

× 24 = 8 oranges

14

of the total oranges = 14

× 24

= 6 oranges

Number of sold oranges = 8 + 6 = 14

Number of left of the oranges = Totalnumber of oranges – Number of soldoranges

= 24 – 14

= 10 oranges.

2 3, 6, 12 2 3 3 3 1

, 3, 6, 3, 3, 1, 1

12 34 cm

4 18 cm

8 12 cm5 3

4 cm

E D

A C

B

4 18 cm

2 2, 4, 8 2 1 2 1 1

, 2, 4, 1, 2, 1, 1


8. Here, 1

56

= 5 6 16

× + =30 1

6+

=316

Now,1

56

÷ 92

=316

÷92

=316

× 29

= 31 26 9××

=31

3 9× =

3127

=4

127

.

9. Initially, Shyam has money = ` 240.

Money spent by shyam 28 part.

Money left with shyam

= 1 – 28

= 68

= 34

part.

Money left with Shyam

= 34

of ` 240

= ` 3 240

4×

= ` 180.

10. 2 8, 16, 24 2 4, 8, 12 2 2, 4, 6 2 1, 2, 3 3 1, 1, 3 1, 1, 1

∴ LCM of 8, 16 and 24= 2 × 2 × 2 × 2 × 3 = 48

∴18

= 1 68 6××

=648

516

= 5 3

16 3××

= 1548

and 724

= 7 224 2×× =

1448

∵ 6, 14, 15 are in ascending order

∴ 648

, 1448

, 1548

are in ascending order

∴18

, 724

, 5

16are in ascending order.

WORKSHEET–13

1. (i) 34

of 32 kg =34

× 32 kg

=3 32

4×

kg = 3 × 8 kg

= 24 kg.

(ii)27

of 1 week =27

of 7 days

(∵ 1 week = 7 days)

=27

× 7 days

= 2 days.

(iii)45

of ` 120 =45

× ` 120

= `4 120

5×

= ` 4 × 24

= ` 96.

2. (i) 234

= 2 4 3

4× +

=114

;

1 13

= 1 3 1

3× +

= 43

Now, 3 1

2 – 14 3

×

34

=11 4

–4 3

×

34

= ( )11 3 – 4 412

× ××

34

= 33 – 16

12

×34

= 17 312 4

××

= 5148

= 1716 = 1

116 .

23CARF IT NO S

(ii)1

34

=3 4 1

4× +

=134

1

23

=2 3 1

3× +

=73

1

14

=1 4 1

4× +

=54

Now, 1

34

×1

23

– 11

4×

15

= 134

×73

–54

×15

= 13 74 3

× –

5 14 5

×

= 13 74 3×× –

5 14 5×× =

9112

–14

= 91 1 – 1 3

12× ×

= 91 – 3

12

= 8812

= 223

= 1

73

.

3. (i) Length of rectangle = l = 4 cm

Breadth of rectangle = b = 1

12

cm

= 1 2 1

2× +

cm

= 32

cm

Area of rectangle = l × b

= 4 cm ×32

cm

=4 3

2×

cm2

= 2 × 3 cm2

= 6 cm2.

(ii) Length of rectangle = l = 39

4cm

=9 4 3

4× +

cm

=394

cm

Breadth of rectangle = b =1

42

cm

=4 2 1

2× +

cm

=92

cm

Area of rectangle = l × b

= 394

cm × 92

cm = 39 94 2×× cm2

=3518

cm2 = 7

438

cm2.

4.1

12 =

1 2 12

× +=

32

and 1

84

=8 4 1

4× +

= 334

∵ Weight of 1 watermelon

= 1

12

kg = 32

kg

∴ Weight of 1

84

watermelons

= 32

×1

84

kg

= 32

×334

kg =3 332 4××

kg

= 998

kg = 3

128

kg.

5.35

of 30 km =35

× 30 km = 3 30

5×

km

=905

km = 18 km.

28 of 40 km =

28

× 40 km = 2 40

8×

km

=808

km = 10 km

Since, 18 km is greater than 10 km

∴ Difference = (18 – 10) km

= 8 km.


6. (i) ∵ 1

23

=2 3 1

3× +

= 6 1

3+

= 73

∴ 8 ÷1

23

= 8 ÷73

= 8 ×37

=8 3

7×

= 247

=3

37

.

(ii)1

84

=8 4 1

4× +

= 32 1

4+

=334

51

6 =

1 6 56

× + =

6 56+

= 116

Now,1

84

÷5

16

= 334

÷116

=334

× 611

=33 64 11××

= 3 32× =

92

= 14

2.

7. Side = 4

45

cm = 4 5 4

5× +

cm

= 20 4

5+

cm = 245

cm

Perimeter = 4 × Side

= 4 ×245

cm = 4 24

5×

cm

= 965

cm = 1

195

cm.

8. Let the man initially had ` a.

Expenditure = 25

of a = 25

× a = 25a

So, money left with him = a – 25a

= 35a

∴ 35a

= 120 or a = 120 5

3×

= 200.

Thus, the man initially had ` 200.

9. (i) 4

18 =

29

; 3520

= 74

Now, 4

18×

3520

×9

14=

29

×74

×9

14

= 2 7 99 4 14× ×× ×

= 14

.

(ii)2

93

=9 3 2

3× +

=293

;

11

29= 1 29 1

29× + =

3029

; 615

=25

.

17

5 = 7 5 1

5× + =

365

Now, 2

93

×1

129

× 615

×1

75

= 293

×3029

× 25

× 365

= 29 30 2 363 29 5 5× × ×× × ×

= 2929

× 305 5×

×363

× 2

= 1 × 65

× 12 × 2

=6 12 2

5× ×

= 144

5 =

428

5.

WORKSHEET–14

1. (i) 57

÷1514

=57

×1415

=5

15×

147

= 13

× 2 = 23

.

(ii)2

65

÷97

=6 5 2

5× +

÷97

= 325

÷97

=325

×79

=32 75 9××

=22445

=44

445

.

2. (i) 8 ×72

=82

× 7 = 4 × 7 = 28.

25CARF IT NO S

(ii) 40 ×58

=408

× 5 = 5 × 5 = 25.

3. (i) 26

13=

2 13 613

× +=

3213

;

11

26=

1 26 126

× +=

2726

Now, 26

13÷

11

26=

3213

÷2726

= 3213

×2627

= 6427

= 21027

.

(ii)3

47

=4 7 3

7× +

=317

;

11

7 =

1 7 17

× + =

87

Now, 3

47

÷1

17

=317

÷87

= 317

×78

=318

×77

= 318

× 1

=318

= 7

38

.

4. 1 hour = 60 minutes= 60 × 60 seconds

58 of 3 hours =

58

× 3 hours

= 58

× 3 × 60 × 60 seconds

= (5 × 3 × 60) × 608

seconds

= 900 ×152 seconds

=9002

× 15 seconds

= 450 × 15 seconds= 6750 seconds.

5. Length of ribbon = 127

2m

=27 2 1

2× +

m

=552

m

Length of 1 piece = 3

24

m =2 4 3

4× +

m

= 114

m

Number of pieces

=Length of ribbonLength of 1 piece

=

552

114

=552

×4

11

=5511

×42

= 5 × 2 = 10 pieces.

6.2

23

= 2 3 2

3× +

= 83

Marks got by Bulbul = 2

23

of mark got

by Kanika = 83

× 75

= 8 ×753 = 8 × 25

= 200 marks.7. First find LCM of 8, 9, 16 and 36

2 8, 9, 16, 36 2 4, 9, 8, 18 2 2, 9, 4, 9 2 1, 9, 2, 9 3 1, 9, 1, 9 3 1, 3, 1, 3 1, 1, 1, 1

∴ LCM of 8, 9, 16 and 36= 2 × 2 × 2 × 2 × 3 × 3 = 144.


We have,

29

= 2 169 16×× =

32144

[∵ 144 ÷ 9 = 16]

18

= 1 188 18×× =

18144

[∵ 144 ÷ 8 = 18]

516

= 5 9

16 9××

= 45

144 [∵ 144 ÷ 16 = 9]

736

= 7 436 4××

= 28

144 [∵ 144 ÷ 36 = 4]

We know that18 < 28 < 32 < 45

∴18144 <

28144 <

32144 <

45144

⇒18 <

736 <

29 <

516 .

8. Total number of students = 50(i) Number of students like playingcricket

=15

of 50 =15

× 50 = 10.

(ii) Number of students like playingfootball

=25

of 50

=25

× 50 = 20

Number of students like playing table-tennis

= 50 – (10 + 20)= 50 – 30 = 20.

(iii) Number of students like playing bothcricket and football = 10 + 20 = 30.

9. (i)1

225

=22 5 1

5× +

=110 1

5+

=1115

12

5 = 2 5 1

5× +

= 10 1

5+

= 115

Now, 1

225

– 12

5 =

1115

–115

= 111 – 11

5

= 100

5 = 20.

(ii) 7–18

= 7 8 – 18

× = 56 – 1

8

= 558

= 7

68

.

10. Number of rotten apples

= 3

10× 1500

= 3 × 150 = 450.

Number of riped apples = 13

× 1500

= 500.❏❏

27ICED AM SL

WORKSHEET – 15

1. (B) ∵ 0.33 > 0.30 > 0.03

∴ 3.33 > 3.30 > 3.03

∴ 3.33 is the greatest.

2. (A) 6 paise = ` 6100

= ` 0.06

6 rupees and 6 paise = ` 6 + ` 0.06= ` 6.06.

3. (A)

4. (D)

5. (C) 7.0683 007.0683100 100

= = 0.070683.

6. (B) 7.752.5

= 7.75 1002.5 100

××

= 775250

= 3110

= 3.1.

7. (A)

8. (D)Perimeter = 3.1 cm + 3.03 cm + 4.2 cm

= (3.1 + 3.03 + 4.2) cm

= 10.33 cm.

9. (A) 4.08 × 100 = 408100

× 100 = 408.

10. (C)

11. (B) ∵ 1 m = 100 cm∴ 0.02 m = 0.02 × 100 cm

= 2

100 × 100 cm = 2 cm.

3Chapter

DECIMALS

12. (B) 11.6 × 0.07 = 11610 ×

7100 =

8121000

= 0.812.

13. (C) 46 ÷ 0.04 = 46 ÷ 4

100 = 46 ×

1004

= 462 ×

1002 = 23 × 50

= 1150.

14. (D) 31.01÷ 0.07 = 3101100 ÷

7100

= 3101100 ×

1007

=3101

7= 443.

15. (A)

16. (A) a × b = 0.72 × 3.03

= 72

100 × 303

100 =

72 30310000

×

= 2181610000

= 2.1816.

17. (A)∵ 1 km = 1000 m = 1000 × 1000 mm

= 1000000 mm

∴ 1 mm = 1

1000000 km

∴ 420 mm = 420

1000000 km

= 00004201000000

= 0.000420 km

= 0.00042 km.

18. (D)

2.38 3.46

5.84+

26.00 18.40

7.6 0

`– `

`

8.000– 3.187 4.813

101.2 km– 88.0 km 13.2 km

89.08– 69.09 19.99

31.46 26.67

58.13+


19. (C) 4 kg 200 g = 4 kg + 200 g

= 4 kg +200

1000kg

= 4 kg + 0.2 kg = 4.2 kg.

7 kg 900 g = 7 kg + 900 g

= 7 kg +900

1000 kg

= 7 kg + 0.9 kg = 7.9 kg.

∴

Now 12.1 kg = 12 kg + 0.1 kg

= 12 kg + 0.1 × 1000 g

= 12 kg + 100 g

= 12 kg 100 g.

20. (B) Area of rectangle

= Length × breadth

= 5.3 × 3.7 = 5310

× 3710

= 53 37

100×

= 1961100

= 19.61 cm2.

21. (C) Perimeter of equilateral triangle= 3 × Side

= 3 × 2.09 = 3 × 209100

= 627100

= 6.27 cm.

22. (B) We know that

0.090 > 0.009

∴ 3.090 > 3.009

Also, 0.777 > 0.704 > 0.007

∴ 2.777 > 2.704 > 2.007

Hence 3.090 > 3.009 > 2.777 > 2.704 >2.007.

23. (A)

WORKSHEET – 16

1. (i) 0.2 × 10 = 2

10 × 10 = 2.

(ii) 4.4 × 10 = 4410 × 10 = 44.

(iii) 3.225 × 10 = 32251000 × 10 =

3225100

= 32.25.

(iv) 0.14 × 100 = 14

100 × 100 = 14.

(v) 3.75 × 100 = 375100

× 100 = 375.

(vi) 8.14 × 100 = 814100

× 100 = 814.

(vii) 1.52 × 1000 = 152100 × 1000

= 152 × 10 = 1520.

(viii) 8.88 × 1000 = 888100

× 1000

= 888 × 10 = 8880.

2. (i) 1.25 × 20 = 125100

× 20 = 125

5 = 25.

(ii) 2.75 × 30 = 275100

× 30 = 275 3

10×

= 82510 = 82.5.

(iii) 8.85 × 40 = 885100 × 40 =

35400100 = 354.

(iv) 6.672 × 300 = 66721000 × 300

= 6672 3

10×

= 20016

10

= 2001.6.

(v) 16.17 × 900 = 1617100

× 900

= 1617 × 9 = 14553.

4.2 kg 7.9 kg

12.1 kg+

0.0077.068

11.898 18.973+

29ICED AM SL

(vi) 3.01 × 1100 =301100 × 1100

= 301 × 11 = 3311.

3. (i) 3.7 × 3 = 3710 × 3 =

11110

= 11.1.

(ii) 4 × 12.75 = 4 × 1275100 =

5100100 = 51.

(iii) 1.2 × 3.1 = 1210

× 3110

= 372100

= 3.72.

4. (i) 328.9 ÷ 10 = 328910

×1

10 =

3289100

= 32.89.

(ii) 728.56 ÷ 10 = 72856

100 × 1

10

= 728561000 = 72.856.

(iii) 0.018 ÷ 10 = 18

1000 × 1

10 =18

10000= 0.0018.

(iv) 0.9257 ÷ 100 = 9257

10000 ×

1100

= 9257

1000000 = 0.009257.

(v) 1.735 ÷ 100 = 17351000 ×

1100

= 1735

100000 = 0.01735.

(vi) 0.02 ÷ 100 = 2

100 × 1

100 = 2

10000= 0.0002.

(vii) 20.2 ÷ 1000 = 20210

× 1

1000

= 202

10000 = 0.0202.

(viii) 2.625 ÷ 1000 = 26251000

× 1

1000

= 2625

1000000 = 0.002625

5. (i) 5.134 ÷ 1.7

= 51341000 ÷

1710

= 51341000 ×

1017

= 513417 ×

1100 =

302100 = 3.02.

(ii) 2.73 ÷ 1.3 = 273100 ÷

1310 =

273100 ×

1013

= 27313 ×

110 =

2110 = 2.1.

6. 2 dozen = 2 × 12 = 24 ∵ Cost of 1 apple = ` 2.50 ∴ Cost of 24 apples = ` 2.50 × 24

= ` 250 24

100×

= ` 6000100 = ` 60.

7. Distance covered by the car = 14.4 kmTime required to cover this distance

= 1.2 hoursSo average distance covered by it in 1hour

= 14.41.2

= 14412

= 12 km.

8.

So 75 km is less than 85.6 km by10.6 km.

9.

4.813 must be added to 4.187 to get 9.

10. 0.25 = 250

1000 and 0.025 = 25

1000

∵ 250 > 25

30217 5134 51 34 34 0

85.6– 75.0 10.6

9.000– 4.187 4.813


∴250

1000 > 25

1000i.e., 0.25 > 0.025So 0.25 is greater.

11. Place value of 2 in 0.321 = 2

100 = 0.02.

WORKSHEET – 17

1. (i) 0.062 × 10 = 62

1000 × 10 = 62

100

= 0.62.

(ii) 73.525 × 100 = 735251000 × 100

= 73525

10 = 7352.5.

(iii) 14.71 × 1000 = 1471100 × 1000

= 1471 × 10 = 14710.

(iv) 0.924 × 100 = 924

1000 × 100 = 92410

= 92.4.

2. (i) 3 × 7.42 = 3 × 742100

= 2226100 = 22.26.

(ii) 1.575 × 8 = 15751000 × 8 =

126001000

= 12.6.

(iii) 8.17 × 300 = 817100 × 300 = 817 × 3

= 2451.

(iv) 7.17 × 600 = 717100 × 600 = 717 × 6

= 4302.

3. (i) 0.04 ÷ 100 = 4

100 × 1

100 = 4

10000= 0.0004.

(ii) 4.47 ÷ 100 = 447100 ×

1100

= 447

10000 = 0.0447.

(iii) 11.5 ÷ 10 = 11510 ×

110 =

115100

= 1.15.

(iv) 0.046 ÷ 1000 = 46

1000 ×

11000

= 46

1000000 = 0.000046.

4. (i) 1.8 ÷ 0.06 = 1810

÷ 6

100 =

1810

× 100

6

= 186 ×

10010 = 3 × 10

= 30.

(ii) 57 ÷ 0.3 = 57 ÷ 3

10 = 57 ×

103

= 573

× 10

= 19 × 10 = 190.

(iii) 11.84 ÷ 0.4 = 1184100

÷ 4

10

=1184100 ×

104 =

11844

×10

100

= 29610 = 29.6.

(iv) 6.6 ÷ 0.11 = 6610 ÷

11100

= 6610 ×

10011

= 6611 ×

10010 = 6 × 10

= 60.

5. (i) ∵ ` 1 = 100 paise

∴ ` 7.25 = 7.25 × 100 = 725 paise.

31ICED AM SL

(ii) ∵ 1 km = 1000 m

∴ 55 km = 55 × 1000

= 55000 metres.

6.

Perimeter of rectangle

= 2 × (length + breadth)

= 2 × (7.25 + 3.15)

= 2 × 10.40 = 20.8 cm.

7. ∵ 1 m = 100 cm

∴ 75 m = 75 × 100 = 7500 cm.

8. 0.02 = 2

100 ; 0.2 = 2

10 = 20

100

∵ 20 > 2 ∴ 20

100 > 2

100

∴ 0.2 > 0.02

Now, 0.2 – 0.02 = 20

100 – 2

100 = 18

100

= 0.18

So, 0.2 is greater than 0.02 by 0.18.

9. 7.25 ÷ 0.5 = 725100

÷ 5

10 = 725100 ×

105

= 725

5 × 10

100 = 14510 = 14.5.

10. Diameter of a circle = 2 × Radius

= 2 × 3.25

= 2 × 325100 =

650100

= 6.5 m.

11. 17.75 × 2.5 = 1775100 ×

2510 =

443751000

= 44.375.

WORKSHEET–18

1. 9.487 ÷ 3.58 = 94871000

÷ 358100

= 94871000 ×

100358

= 9487358

× 100

1000

= 26.510 = 2.65.

2. (i) 8.08 × 1000 = 808100 × 1000

= 808 × 10 = 8080.

(ii) 0.96 ÷ 100 = 96

100 × 1

100

= 96

10000 = 0.0096.

3. 1.2 × 1.2 = 1210 ×

1210 =

12 1210 10

××

= 144100 = 1.44.

4. 3 dozens = 3 × 12 = 36

∵ Cost of 1 banana = ` 1.25

∴Cost of 36 bananas = ` 1.25 × 36

= ` 125100 × 36

= ` 4500100 = ` 45.

5. Required number = 1.4

0.014 = 1.4000.014

= 1400

14 = 100.

6.25 paise

1` =

25 paise100 paise

= 14

= 0.25

So 25 paise is 0.25 part of a rupee.

7. Quantity of oil for one dish

= Total quantity of oil

No. of dishes

7.25 3.15

10.40+


= 3.204

9 =

32041000

× 19

= 356

1000 = 0.356 litre

= 0.356 × 1000 ml

= 356 ml.

8. Total quantity of fruits = 5 kg

Quantity of fruits consumed by parents= 2.5 kg

Quantity of fruits consumed bychildren = 1.25 kg

So, quantity of fruits consumed by thefamily = 2.5 + 1.25

= 3.75 kg

Quantity of remaining fruits

= 5 – 3.75

= 1.25 kg

9. 0.99 = 99100

; 0.09 = 9

100; 0.90 =

90100

∵ 99 > 90 > 9

∴ 99

100 > 90

100 > 9

100or 0.99 > 0.90 > 0.09

i.e., 0.99 is the greatest.

10. The perimeter of a triangle is the sumof its sides.

Perimeter of equilateral triangle

= 3 × Length of one side

= 3 × 1.5 = 3 × 1510

= 4510

= 4.5 cm.

11. Area of square = Side2

= ( 2.5)2 = 2.5 × 2.5

= 2510

× 2510

= 625100

= 6.25 cm2.

12. ∵ Cost of 1 pair of shoes = ` 179.50

∴Cost of 3 pairs of shoes

= ` 179.50 × 3

= ` 538.50

∵ Cost of 1 pair of sandals = ` 216.25

∴Cost of 4 pairs of sandals

= ` 216.25 × 4

= ` 865.00

Clearly, cost of 4 pairs of sandals ismore than the cost of 3 pairs of shoes.

13. (i) 7.750.25

= 77531

25= .

(ii) 42.80.02

= 42.800.02

= 4280

2 = 2140.

WORKSHEET – 19

1. (i) 25.5 ÷ 5 = 25510

÷ 5

= 25510

× 15

= 2555

× 1

10

= 51 × 1

10 = 5.1.

(ii) 126.35 ÷ 7 = 12635

100 ÷ 7

= 12635

7 ×

1100

= 1805100

= 18.05.

2. (i) 6 km = 6 × 1000 m = 6000 m.

(ii) 700 m = 7001000

km = 710

km

= 0.7 km.

(iii) 17956 g = 179561000

kg = 17.956 kg.

2.50+ 1.25 3.75

5.00– 3.75 1.25

17950 3

53850×

21625 4

86500×

18057 12635 7

56 5635 35

0

33ICED AM SL

3. Capacity of 1 bucket = 16.35 litres

= 1635100 litres

Capacity of 12 buckets

= 1635100 × 12 litres

= 19620

100 litres

= 196.20 litres.

4. (i) 5.78 × 3 = 578100 × 3

= 578 3

100×

= 1734100 = 17.34.

(ii) 7.248 × 0.19 = 72481000 ×

19100

= 7248 19100000

×

= 137712100000

= 1.37712.

(iii) 7.248 × 400 = 72481000 × 400

= 724810 × 4

= 7248 4

10×

= 28992

10= 2899.2.

5. (i) 37.17 ÷ 7 = 3717100 ÷ 7

= 3717100

× 17

= 3717

7 ×

1100

= 531100 = 5.31.

(ii) 15.064 ÷ 28

= 150641000

÷ 28

= 150641000

×1

28

=15064

28×

11000

= 538

1000= 0.538.

6. Monthly expenditure

= 0.75 of 12000

= 0.75 × 12000

= 75

100 × 12000 = 75 × 120

= ` 9000.

Monthly saving = Salary – Expenditure= 12000 – 9000 = ` 3000

Number of months = 39000

Monthly saving

= 390003000

= 393

= 13.

7. Other number

= Product

One number

= 136.369

2.65

= 136369

2650= 51.46.

8. Perimeter of a regular polygon= No. of sides × Length

of one side

⇒ 22.5 = No. of sides × 2.5

⇒ No. of sides = 22.52.5 =

22525 = 9.

9. Thickness = 19.15 mm

= 19.15

10 cm

[∵ 1 cm = 10 mm]

= 1.915 cm.

1635 12

3270163519620

×

×

7248 19

652327248137712

×

×

7248 4

28992×

5317 3717 35

21 21 7 7

0

106 84 224 224

0

120 75600

8409000

×

×

51.462650 136369 13250

3869 265012190 1060015900 15900

0

53828 15064 140


WORKSHEET – 20

1. Length of each piece

= Length of ribbonNumber of pieces

= 19.63

13 = 1963100 ×

113

= 1963

13 × 1

100 = 151100 = 1.51 m.

2. (i) 0.018 ÷ 0.13 = 0.0180.13

= 0.0180.130

= 18

130

= 0.13846.

(ii) 13.455 ÷ 4.1

= 13.455

4.1

= 13.4554.100

= 134554100

= 3.2817

(iii) 441.709 ÷ 18

= 441.709

18

= 441.70918.000

= 44170918000

= 24.539388

= 24.53939.

(iv) 1.001 ÷ 7 = 1.001

7 =

10011000

× 17

= 1001

7 ×

11000

= 143

1000

= 0.143.

3. Area of rectangle

= Length × Breadth

= 12.5 × 8.3

= 12510

× 8310

= 10375

100 = 103.75 cm2.

4. Sum of costs of 1 toy and 1 box

= ` 106.35 + ` 18.65

= ` 125

No. of pairs of toys and boxes

= 1000 125

``

= 1000125

= 8

So, 8 toys and 8 colour boxes can bebought.

5. (i) 7 m = 7

1000km = 0.007 km.

(ii) 9 m = 9 × 100 cm = 900 cm.

(iii) 7.3 km = 7.3 × 1000 m = 7300 m.

(iv) 0.055 kg = 0.055 × 1000 g = 55 g.

(v) 6 kg 5 g = 6 × 1000 g + 5 g

= 6000 g + 5 g = 6005 g.

15113 1963

13 66 65 13 13 0

0.138461130 180 130

500 3901100 1040 600 520800 780200

130 70

3.28174100 13455 12300

11550 820033500 32800 7000 4100 29000 28700

300

12583375

100010375

×

×

24.53938818000 441709 36000

81709

70000 54000160 000144000160000 144000

16000

72000 97090 9000070900 54000169000 162000 70000

106.3518.65

125.00+

35ICED AM SL

(vi) 3 m 55 cm = 3 × 100 cm + 55 cm

= 300 cm + 55 cm

= 355 cm.

(vii) ` 9.25 = 9.25 × 100 paise

= 925 paise.

(viii) 5580 paise = ` 5580100

= ` 55.8.

6. (i) Place value of 6 in 8.36 = 0.06.

(ii) Place value of 4 in 12.294 = 0.004.

7. Sum of given numbers

= 15.29 + 11.729

= 27.019

Difference of given numbers

= 15.29 – 11.729

= 3.561

∴ Required difference

= 27.019 – 3.561

= 23.458.

8. Difference of 7.124 and 5.62

= 7.124 – 5.62

= 1.504

Required value

= 10 – 1.504

= 10.000 – 1.504

= 8.496.

9. (i) 1.79 = 179100

; 1.9 = 190100

As 190 > 179, 1.9 is greater than1.79.

(ii) 1.05 = 105100

; 1.50 = 150100


(iii) 0.8 = 80

100; 0.88 =

88100

As 88 > 80, 0.88 is greater than 0.8.

(iv) 3.33 = 333100 ; 3.30 =

330100


WORKSHEET – 21

1. ∵ Cost of 6 pens = ` 18.36

∴ Cost of 1 pen = ` 18.36

6

∴ Cost of 10 pens = ` 18.36

6× 10

= ` 1836600 × 10

= ` 1836

6 ×

110

= ` 30610

= ` 30.60.

2. 2 weeks = 2 × 7 = 14 days

∵ No. of pages typed in 1 day = 10.50

∴ No. of pages typed in 14 days

= 10.50 × 14

= 1050 14

100×

= 14700

100 = 147.

3. Average of 1.3, 3.2, 4.5 and 5.8

= 1.3 3.2 4.5 5.8

4+ + +

= 14.84

=148

4 × 1

10

= 3710

= 3.7.

4. 7.232 = 72321000

; 7.322 = 73221000

∵ 7322 > 7232 ∴ 73221000

> 72321000

So 7.322 is greater.

15.29011.729

27.019+

15.290–11.729 3.561

27.019– 3.56123.458

7.124– 5.620 1.504

10.000– 1.504 8.496

1050 14 4200105014700

×

×


5. Let a would be added

∴ a + 1.35 = 3

⇒ a = 3 – 1.35 (Transposing)

⇒ a = 3.00 – 1.35

= 1.65.

6. ∵ 1 year = 12 months

∴ 2 years = 2 × 12 = 24 months

∵ Weight of rice consumed in 1 month= 12.5 kg

∴ Weight of rice consumed in 24months = 12.5 × 24 kg

= 125 24

10×

kg

= 300010

kg = 300 kg.

7. (i) 2.02 × 1000 = 202100

× 1000

= 202 × 10 = 2020.

(ii) 0.52 ÷ 100 = 52

100 ×

1100

= 52

10000 =

0005210000

= 0.0052.

8. (i) 11 cm = 11

100 m = 0.11 m.

(ii) 63 kg = 63 × 1000 g = 63000 g.

9. ∵ Price of 1 kg of wheat = ` 12

∴ Price of 43.7 kg of wheat

= ` 12 × 43.7

= ` 12 43710×

= ` 524410

= ` 524.4.

Amount of money spent for each

person = ` 524.4152

= ` 5244152

× 1

10

= ` 34.510

= ` 3.45.

10. 175 km is less than 385.6 km by thedifference of them.∴ Required value

= 385.6 – 175= 385.6 – 175.0 = 210.6 km .

11. Total weight bought by Varsha= 7 kg 300 g + 2 kg 500 g= 7000 g + 300 g + 2000 g

+ 500 g(∵ 1 kg = 1000 g)

= 9800 gTotal weight bought by Reema

= 5 kg 800 g + 3 kg 100 g= 5000 g + 800 g + 3000 g + 100 g = 8900 g

∵ 9800 > 8900∴ Varsha bought more rice and wheataltogether.

12. (i) 185.5 ÷5 = 185.5

5 =

185550

= 37110

[Dividing by 5]

= 37.1.

(ii) 186.55 ÷ 7 = 186.55

7

= 18655

100 ×

17

= 18655

7 ×

1100

= 2665100

= 26.65.❑❑

43712

8744375244

×

×

34.5152 5244 456

684 608 760

760 0

385.6–175.0210.6

7000300

2000 500

9800+

5000800

3000 100

8900+

3.00– 1.351.65

12524

5002503000

×

×

26657 18655 14

46 42 45 42 35 35

0

37ATA H A DN L NI GD

WORKSHEET – 221. (D) Range = greatest observation

– smallest observation

= 8 – 2 = 6.

2. (C) Mean age (in years)

= 11 11.5 12 14 12.5

5+ + + +

= 615

= 12.2.

3. (B) Mean = 1 2 3 4 5 6

6+ + + + +

= 216

= 3.5.

4. (B) Arithmetic mean = 0 1 2 3 4

5+ + + +

= 105

= 2.

5. (C)As 6 occurs maximum number oftimes, 6 is the mode.

6. (B) The ascending order of the data: 2, 3, 4, 5, 7, 8, 9Since the middle most term is 5, so themedian is 5.

7. (A) Since, 4 is used maximum numberof times, so 4 is the mode.

Range = 6 – 1 = 5.8. (C) Mean of 4, 6, 8 and 14

= 4 6 8 14

4+ + +

= 324

= 8

Mean of 6, 8, 12 and 14

= 6 8 12 14

4+ + +

= 404

= 10

Now, mean of 8 and 10 = 8 10

2+

= 182

= 9.

4Chapter

DATA HANDLING

9. (B) Hint: {H, T}.10. (D) Probability of getting a head

= No. of heads

No. of all possible outcomes

= 12

.

11. (A) There is only one card marked with3.

∴ Required probability = 15

.

12. (B) Probability of a sure event isalways 1.

13. (B) Middle most term = 5th term = 2

∴ Median = 2.14. (D) Range = greatest observation

– smallest observation= 12.2 – 0.0 = 12.2 mm.

15. (B) The greatest observation is 142 cm

∴ Height of the tallest girl = 142 cm.16. (C) Only 22 repeats in the given data

∴ Mode = 22 runs.17. (A) A die has 4 vertical faces and 2

horizontal faces.18. (B) Mean

= 11 12 13 14 15 16 17 188

+ + + + + + +

= 116

8 = 14.5.

19. (A) A die has no 7 as marked number∴ Getting a 7 is an impossible event∴ Probability = 0.


WORKSHEET – 23

1. Greatest observation = 10 yearsSmallest observation = 5 years∵ Range = 10 – 5 = 5 years.

2. (i) Amit is the heaviest.Ramu is the lightest.

(ii) Sum of the weights = 77 + 58 + 62 + 81 + 73

= 351 kg.Mean of the weights

= Sum of the weightsNumber of students

= 3515

= 70.2 kg .

3. Since 2 occurs maximum number oftimes.So, the mode is 2.

4. Arrange the given data in ascendingorder. 25, 30, 30, 30, 40, 45, 50, 55, 60, 60(i) Since 30 occurs maximum number

of timesSo, 30 runs is the mode.

(ii) Number of observations, n = 10Since n is even

∴ Median = 12 ( )th

2n

observation

+ ( )1 th observation2n +

=12

[5th observation + 6th

observation]

= 12

[40 + 45] = 12

× 85 = 42.5 runs.

5. First arrange the given data indescending order as given below: 21, 19, 17, 14, 13, 13, 13, 11

(i) Range = Greatest observation – smallest observation

= 21 – 11 = 10.(ii) Sum of the observations

= 21 + 19 + 17 + 14 + 13 + 13 + 13 + 11= 121Number of the observations = 8

Mean = Sum of the observationsNo. of the observations

= 1218 = 15.125.

(iii) Number of observations, n = 8which is even

∴ Median =12 ( )th observatioin

2n


= ( )1 8th

2 2 observation

+ ( )81 th observatiion

2+

= 1

[4th observation2

5th observation]+

= 12 [14 + 13]

= 12 × 27 = 13.5.

(iv) Mode = The observation whichoccurs maximum numberof times

= 13. 6.(i) If H represent a head and T a tail,

then the sample space is given by:S = {H, T}

Now, probability of getting a head = 12 .


(ii) When a die is thrown once, thesample space is given by: S = {1, 2, 3, 4, 5, 6}

Now probability of getting '2' = 16 .

(iii) Probability of choosing a girl

= Number of girls

Sum of boys and girls

= 2

4 2+ =

26

= 13

.

7. (i) Number of marbles marked 3 = 1Probability of drawing a marble

marked 3 = 16

.

(ii) Number of marbles marked 6 = 1Probability of drawing a marble

marked 6 = 16

.

8. (i) Number of outcomes

= Number of letters in the word'SPINNING'

= 8.(ii) A letter who occurs maximumnumber of times, has the highestprobability.As, 'N' occurs maximum number oftimes, i.e., 3 times, 'N' has highestprobability.

∴ P(N) = 38

.

(iii) Probability for letter 'I'

= Number of times occurring 'I'

Number of outcomes

= 28

= 14

. [Using part (i)]

9. To draw a double bar graph, you haveto go to the following steps:Step I. Draw a pair of perpendicularlines OX and OY on a graph paper.Step II. Along the horizontal axis (OX),mark the names of students, namelyNavin, Anuj, Poonam, Meera andKiran.


Along the vertical axis (OY), mark themarks obtained by the students.Step III. Choose a suitable scale todetermine the height of bars. Here, take1 mark = 4 small divisions on thegraph.Step IV. First draw the bars for II termand then for III term for differentstudents.Bars for II term and III term are shadedseparately and their shadings areshown in the top right corner of thegraph paper. Write the marks of theevery term on the top of correspondingbar.

WORKSHEET–24

1. (i) Arranging the given observationsin the ascending order, we get 2, 2, 3, 3, 3, 4, 4, 5, 5Range = greatest observation

– smallest observation = 5 – 2 = 3

Mean = Sum of observations

Number of observations

= 2 2 3 3 3 4 4 5 5

9+ + + + + + + +

= 319

= 3.44

Median = Middle most term

= ( )9 1th

2+

term

= 5th term = 3Mode = observation that occurs

most often= 3.

(ii) Arranging the given observationsin the ascending order, we get 10, 10, 11, 11, 12, 12, 15, 15, 15

Range = Greatest observation – smallest observation

= 15 – 10 = 5

Mean = Sum of observationsNumber of observations

= 10 10 11 11 12 12 15 15 15

9+ + + + + + + +

= 111

9 = 12.33

Median = Middle most term

= ( )9 1th

2+

observation

= 5th term = 12 Mode = observation that occurs

most often= 15.

2. Arranging the given observations inthe ascending order, we have 8 m,12 m, 14 m, 14 m, 16 m, 20 m, 24 mn = 7; which is odd

Median = ( )7 1th

2+

term

= 4th term = 14 m.3. Mean temperature

= Sum of all observationsNumber of observations

= 21 23 25 24 22 24 23 25 25 21

10+ + + + + + + + +

= 23310

= 23.3°C.

4. Total number of members in the group = 7 + 8 + 3 = 18

Number of children = 3Probability of selecting a child

= Number of childrenTotal number of members

= 318

= 16

.


5.∵ Number of 3's = 1∴ Number of favourable outcomes

= 1Number of all possible outcomes = 6

∴ Probability of getting '3' = 16 .

6. Let us put the given data in a tabularform:

Numbers Tally Marks No.of matches1 ||| 32 |||| 43 | 14 || 2

From the table, it is clear that the modeis 2.

7. In order to construct a bar graph, youhave to go to the following steps:Step I. Take a graph paper and draw apair of perpendicular lines OX and OY.Call OX as the horizontal axis and OYas the vertical axis.Step II. Along OX, mark the names of

colours and choose the equal width ofthe bars and uniform gap betweenthem.Along OY, mark the number ofstudents.Step III. Choose a suitable scale todetermine heights of the bars. You canchoose

1 big division = 5 studentsStep IV. Calculation for heights ofvarious bars:

Height of the bar of pink colour = 425

= 8.4 big divisions= 8 big divisions and 4 small divisions

Height of the bar of red colour = 455

= 9 big divisionsHeight of the bar of blue colour

= 505 = 10 big divisions

Height of the bar of yellow colour


Height of the bar of green colour


Step V. Draw the bars with heightsobtained in step IV and write thecorresponding number of students onthe top of each bar.

8. Arranging the given observations inthe descending order as follows: 42, 38, 35, 34, 32, 32, 32As 42 is not as a middle term, thegiven median is not correct.

Correct median = ( )7 1th

2+

term

= 4th term = 34Since, 32 has highest frequency, sogiven mode is correct.


9.9.9.9.9. Putting the given data in tabular form,we get

Numbers Tally marks Frequency

1 |||| ||| 82 |||| |||| |||| 143 |||| || 74 |||| 55 ||| 36 || 2

Mode:Mode:Mode:Mode:Mode: The highest frequency is of 2.So, 2 is the mode.Median:Median:Median:Median:Median: Let us arrange the given datain ascending order.1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2,2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4,5, 5, 5, 6, 6.Number of observations, n = 39

n is odd, so, median = ( )39 1th

2+

term

= 20th term = 2.

WORKSHEET –25

1. 1. 1. 1. 1. (i) Range = Highest height – lowestheight

= 180 cm – 165 cm

= 15 cm.

(ii) 9 + 8 + 12 = 29 girls have morethan 165 cm of height.

(iii) Sum of heights of all girls

= 9 × 170 + 8 × 175 + 11 × 165

+ 12 × 180

= 1530 + 1400 + 1815 + 2160

= 6905 cm

Number of girls = 40

Mean height = Sum of heights of all girls

Number of girls

= 6905

40 = 1381

8

= 172.625 cm.2.2.2.2.2. Total score = 23 + 60 + 75 + 81 + 55

+ 50 + 70 + 45 + 50 + 90 + 0= 599

Number of members = 11

Mean = Total score

Number of members

= 59911

= 54.45 runs (approx.).

Rearrange the observations inascending order.0, 23, 45, 50, 50, 55, 60, 70, 75, 81, 90Number of observations = 11, which isodd.

Middle most term = ( )1th

2n +

= ( )11 1th

2+

= 6th = 55

∴ Median = 55 runs.3.3.3.3.3. Let us put the given data in tabular

form:

Numbers Tally Marks Frequency12 || 213 || 214 ||| 316 | 119 | 1

The frequency is highest for 14, i.e., 3.So the mode is 14.

4.4.4.4.4. To draw a double bar graph, you haveto go to the following steps:Step I.Step I.Step I.Step I.Step I. Draw a pair of perpendicularlines OX and OY on a graph paper.Step II.Step II.Step II.Step II.Step II. Along the horizontal axis (OX),mark the test numbers, namely I, II,III, IV and V.


Along the vertical axis (OY), mark theaverage marks.Step III. Choose a suitable scale todetermine the height of bars. Here, take

1 mark = 1 small division on thegraph.

Step IV. First draw the bars for theclass VII A and then for VII B takingequal width of the bars and equal gapbetween any two consecutive bar pairs.Shade the bars of the classes withdifferent types. Show their shadingson the top right corner of the graphpaper.

5. To draw a double bar graph, you haveto go to the following steps:Step I. Draw a pair of perpendicularlines OX and OY on a graph paper.Step II. Along the horizontal axis (OX),mark the years and along the verticalaxis (OY), mark the units of books soldof different parts.Step III. Choose a suitable scale todetermine the height of each bar onthe graph paper.

Suppose 50 units = 1 big divisionStep IV. Calculation for heights ofvarious bars:

50 units = 1 big divisionHeight of the bar for part I year 2004

= 45050

= 9 big divisions

Height of the bar for part I year 2005

= 40050

= 8 big divisions


= 63050

= 12.6 big divisions

= 12 big divisions and 6small divisions


= 40050

= 8 big divisions

Height of the bar for part II year 2004

= 50050

= 10 big divisions


= 60050

= 12 big divisions


= 65050

= 13 big divisions



= 53050

= 10.6 big divisions

= 10 big divisions and 6small divisions

Step V.Step V.Step V.Step V.Step V. First draw the bars for the partI and then for the part II taking equalwidth of bars and equal gap betweenany two consecutive combined pairs ofthe bars. Shade the bars for part I inone way and the bars for part II in theother way. Show their shadings in thetop right corner of the graph paper.

6.6.6.6.6. Total number of cards = 52 Number of aces = 4Chance of getting an ace

= Number of aces

Total number of cards

= 4

52 = 1

13.

7.7.7.7.7. All possible outcomes are: 1, 2, 3, 4, 5and 6∴ Number of all possible outcomes = 6

Since favourable outcome is 5∴ Number of favourable outcomes = 1Now, chance of getting 5.

= Number of favourable outcomesNumber of all possible outcomes

= 16

.

8.8.8.8.8. In order to draw a bar graph, you haveto go to the following steps:Step I.Step I.Step I.Step I.Step I. Take a graph paper and draw apair of perpendicular lines OX and OY.Call OX as the horizontal axis and OYas the vertical axis.Step II.Step II.Step II.Step II.Step II. Along OX, mark the names ofthe favourite snacks and along OY,mark the number of students.Step III.Step III.Step III.Step III.Step III. Choose a suitable scale todetermine height of each bar on thegraph paper

Suppose 1 big division = 10 studentsStep IV.Step IV.Step IV.Step IV.Step IV. Calculation for heights ofCalculation for heights ofCalculation for heights ofCalculation for heights ofCalculation for heights ofvarious bars:various bars:various bars:various bars:various bars:∵ 1 big division = 10 students


∴ 1 small divisions = 1 student

Height of bar for Burger = 4310

= 4.3 big divisions= 4 big divisions and 3 small

divisions

Height of bar for Finger chips = 1910

= 1.9 big divisions= 1 big divisions and 9 small

divisions

Height of bar for Pizza = 5510

= 5.5 big

divisions

Height of bar for Sandwich = 4910

= 4.9 big divisions= 4 big divisions and 9 small divisions

Height of bar for Pakora = 3410

= 3.4 big

divisions= 3 big divisions and 4 small divisions

Step V.Step V.Step V.Step V.Step V. Draw the bars of heightsobtained in step IV and of equal widthand equal gap between any twoconsecutive bars.

You can write the number of studentson the top of the corresponding bars.

(i) As the bar for Pizza is the highest,Pizza is the most preferred snack. Asthe bar for finger chips is the shortest,Finger chips is the least preferredsnack.

(ii) There are 5 items (snacks) in all.These are Burger, Finger chips, Pizza,Sandwich and Pakora.

WORKSHEET –26

1. 1. 1. 1. 1. (i) 1, 7, 13, 19, 25, 13, 13Let us represent the data in the tabularform:Numbers Tally Marks Frequency

1 | 17 | 1

13 ||| 319 | 125 | 1

From the table, the frequency of 13 isthe highest, so 13 is the mode.

(ii) 4, 6, 8, 4, 10, 4, 6, 12, 6, 10, 6

Let us represent the data in the tabularform:

Numbers Tally Marks Frequency4 ||| 36 |||| 48 | 1

10 || 212 | 1

From the table the frequency of 6 isthe highest.So, 6 is the mode.

(iii) 26, 32, 26, 21, 83, 26, 83, 67, 53, 26,85.


Let us represent the data in the tabularform:

Numbers Tally Marks Frequency

26 |||| 432 | 121 | 183 || 267 | 153 | 185 | 1

From the table, frequency of thenumber 26 is the highestSo, 26 is the mode.

2. (i) 2, 3, 5, 7, 9These numbers are in ascendingorder.Number of observations, n = 5Since, 5 is odd

So, Median = ( )5 1th

2+

observation

= 3rd observation= 5.

(ii) 60, 33, 63, 61, 44, 48, 57Arranging the observations inascending order, we have33, 44, 48, 57, 60, 61, 63Number of observations, n = 7Since 7 is odd

∴ Median = ( )7 1th

2+

observation

= 4th observation= 57.

(iii) 13, 22, 25, 8, 11, 19, 17, 31, 16, 10Arranging the observations inascending order, we have8, 10, 11, 13, 16, 17, 19, 22, 25, 31Number of observations, n = 10Since 10 is even, so median will be

the mean of ( )th2n

observation and

( )1 th2n + observation.

Here 2n

= 102 = 5 and

2n

+ 1

= 5 + 1 = 6

Now,median = ( )1th observation

2 2n


= 12 [5th observation + 6th observation]

= 12 [16 + 17] =

12 × 33 = 16.50.

3. Sum of heights of all students= 150 × 3 + 151 × 12 + 152 × 9 + 153 ×

6 + 154 × 15 + 155 × 5= 450 + 1812 + 1368 + 918 + 2310 + 775= 7633 cmNumber of students = 3 + 12 + 9 + 6

+ 15 + 5= 50

Mean height = Sum of heights

Number of students

= 7633

50 = 152.66 cm.

4. (i) First 5 natural numbers. are:1, 2, 3, 4 and 5

∴ Mean =1 2 3 4 5

5+ + + +

=155 = 3.

(ii) First 5 prime numbers are:2, 3, 5, 7, 11

∴ Mean =2 3 5 7 11

5+ + + +

=285 = 5.6.

(iii) ` 8 + ` 18 + ` 31 + ` 43 + ` 70= ` (8 + 18 + 31 + 43 + 70)= ` 170

Mean = 1705

` = ` 34.


5.5.5.5.5. Total number of hours

= 1

34 +

12

2 + 3

24

= 134 +

52 +

114

= 134

+ 104

+ 114

= 344 =

172

Number of days = 3

∴ Mean =

1723

=

17231

= 172 ×

13

= 176 =

52

6 hours.

6.6.6.6.6. In order to draw a double bar graph,you have to go to the following steps:Step I.Step I.Step I.Step I.Step I. Draw a pair of perpendicularlines OX and OY on a graph paper.Step II.Step II.Step II.Step II.Step II. Along the horizontal axis (OX),mark the given months.Step III.Step III.Step III.Step III.Step III. Along the vertical axis (OY),you have to mark the rainfall incentimetres. For this, choose anappropriate scale keeping in view themaximum and minimum observations(9 cm and 2 cm).Suppose 1 cm = 1 big divisionStep IV.Step IV.Step IV.Step IV.Step IV. First draw the bars for theyear 1993 and then for the year 1994taking equal width of bars and equalgap between any two consecutivecombined pairs of bars.

Step V.Step V.Step V.Step V.Step V. Shade the bars of both theyears with different colours and showtheir shadings at the top right cornerof the graph paper.

WORKSHEET –27

1.1.1.1.1. In order to draw a bar graph, you haveto go to the following steps:Step I.Step I.Step I.Step I.Step I. Take a graph paper and draw apair of perpendicular lines OX and OYon it. Call OX as the horizontal axis

and OY as the vertical axis.Step II.Step II.Step II.Step II.Step II. Along OX, mark the names ofthe brands.Step III.Step III.Step III.Step III.Step III. Along OY, mark the numberof units sold by taking an appropriatescale keeping in view the minimumand maximum units sold (150 and 260).Suppose 20 units = 1 big divisionStep IV.Step IV.Step IV.Step IV.Step IV. Calculations for heights ofvarious bars:∵ 1 big division = 20 units


∴ 1 small division = 2010 = 2 units

∴ Height of bar for L.G. = 18020

= 9 big divisions

Height of bar for Samsung = 20020

= 10 big divisions

Height of bar for Sony = 26020

= 13 big divisions

Height of bar for Videocon = 15020

= 7.5 big divisions= 7 big divisions

and 5 small divisions

Height of bar for phillips = 25020

= 12.5 big divisioins= 12 big divisions and 5 small divisions

Height of bar for Sansui = 16020

= 8 big divisions

Step V. Draw the bars of heightsobtained in step IV and of equal widthand equal gap between any twoconsecutive bars.You can mention the number of unitssold on the top of corresponding bar.

2. (i) Height of the bar for more than 80marks= Height of the bar for 85 marks= 3 students∴ Required number of students = 3

(ii) The highest bar corresponds to 75marks.So 75 marks were obtained by mostnumber of students

(iii) Number of failed students= Number of students obtaining

to 50 marks + Number ofstudents obtaining to 55marks

= 1 + 1 = 2.(iv) Frequency Table:

Marks (x) Frequency50 155 160 365 470 375 580 185 3

21

Total marks obtained by the students

= 50 × 1 + 55 × 1 + 60 × 3 + 65 × 4+ 70 × 3 + 75 × 5 + 80 × 1 + 85 × 3

= 50 + 55 + 180 + 260 + 210 + 375

+ 80 + 255

= 1465


Mean

=Total marks obtained by the students

Number of students

= 1465

21 = 69.76 marks.

3. (i) Probability (black queen)

= Number of black queens

Total number of cards =

252

= 1

26 .

(ii) Probability (head)


= 12 .

4. (i) Probability (getting 3)


= 16 .

(ii) Probability (getting less than 3)

= 26 =

13 .

(iii) Probability (getting an even no.)

= 36 =

12 .

(iv) Probability (getting 5) = 16 .

5. Arranging the given salaries in thedescending order, we have

` 121, ` 98, ` 89, ` 72, ` 70, ` 70, ` 50,

` 38

Here, number of terms, n = 8, which iseven

Middle terms are ( )8th

2

= 4th and ( )81 th

2+ = 5th

Now, median salary

= 4th term + 5th term

2

= ` 72 70

2+

= ` 142

2 = ` 71.

6. (i) Let us represent the given data inthe ascending order. 3, 3, 3, 3, 5, 7, 7, 8, 9, 9Looking these numbers., we easilycan say that '3' is used maximumnumber of times.So, 3 is the mode.

(ii) Let us arrange the given data inthe ascending order.10, 11, 13, 17, 18, 19, 20, 23, 25, 29,29, 29, 29, 30, 35Looking these numbers, we easilycan say that '29' is used maximumnumber of times.So, 29 is the mode.

7. Total of multiples of f and x= 7 × 5 + 8 × 16 + 20 × 25 + 10 × 35 + 12 × 45= 35 + 128 + 500 + 350 + 540= 1553

Sum of f ’s = 7 + 8 + 20 + 10 + 12 = 57

Now, mean = 1553

57 = 27.25 (approx.).

8. First ten odd natural numbers are: 1, 3, 5, 7, 9, 11, 13, 15, 17, 19Sum of these numbers

= 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17+ 19

= 100

Mean = Sum of numbers

Number of numbers = 10010

= 10.9. First 7 whole numbers are:

0, 1, 2, 3, 4, 5, 6


Sum of these numbers= 0 + 1 + 2 + 3 + 4 + 5 + 6 = 21

Mean = Sum of numbers

Number of numbers

= 217 = 3.

10. Let us arrange the given data in thedescending order. 45, 40, 40, 39, 35, 32, 30, 28, 27Clearly, 45 is the highest observationand 27 is the lowest∴ Range = 45 – 27 = 18.

WORKSHEET – 28

1. Frequency table:

Score Tally marks Frequency4 || 26 |||| 51 | 13 || 27 ||| 39 ||| 35 |||| 48 ||| 3

10 || 22 || 2

Range = Greatest observation – Lowestobservation

= 10 – 1 = 9.2. Mean temperature

= Sum of observations

Number of observations

= 39 37 38 28 30 35 36

7+ + + + + +

= 2437

= 34.71°C (approximately).3. Let us represent the ages in the

ascending order. 25, 27, 28, 32, 36, 38, 40, 41, 54, 57

(i) Age of the oldest teacher is 57 yearsAge of the youngest teacher is 25 years.(ii) Range = (57 – 25) years = 32 years.(iii) Sum of ages

= (25 + 27 + 28 + 32 + 36 + 38 + 40 + 41 + 54

+ 57) years= 378 years

Mean age = Sum of ages

Number of teachers

= 37810

years = 37.80 years.

4. Arranging the given observations inthe descending order, we have 63, 61, 60, 51, 48, 44, 33Number of observations, n = 7, whichis odd

Here, 1

2n +

= 7 1

2+

= 4

Median = 4th term = 51.5. Arranging the given observations in

the ascending order, we have1, 2, 3, 4, 4, 4, 5, 5, 5, 5, 6, 6, 7, 7, 8, 8, 9,10Since, 5 occurs maximum number oftimes, so 5 is the mode.Sum of the observations

= 1 + 2 + 3 + 4 + 4 + 4 + 5 + 5 + 5 + 5 + 6 + 6 + 7 + 7 + 8 + 8 + 9 + 10= 99

No. of observations = 18

Mean = Sum of the observationsNumber of observations

= 9918

= 112

= 5.5.

6. (i) Lata scored the highest marks of 80in English.

(ii) Her lowest score is 30 marks inScience.


(iii) 10 marks = 1 unit.

7. Let us arrange the given outcomes inthe descending order.

6, 6, 5, 5, 5, 4, 4, 4, 3, 3, 3, 2, 2, 1, 1

(i) As 5 occurs 3 times, frequency of 5is 3.

(ii) As 1 occurs 2 times, frequency of 1is 2.

8. In order to draw a bar graph, you haveto go to the following steps:

Step I. Take a graph paper and draw apair of perpendicular lines OX andOY. Call OX as the horizontal axis andOY as the vertical axis.

Step II. Along OY mark the observat-ions by taking an appropriate scalekeeping in view the minimum andmaximum observations (10 and 260).

Suppose 20 = 1 big division

Step III. On OX, you have to draw thebars of equal width.

Let us determine their heights.

Calculation for heights of various bars:

Height of bar for 130 = 13020

= 6.5 big divisions.


= 2 big divisions


= 0.5 big divisions


= 1 big divisions


= 13 big divisions


= 5.5 big divisions


= 1.5 big divisions


= 4.5 big divisions.

Step IV. Draw the bars along OX usingtheir heights obtained in step III. Thegap between any two pairs ofconsecutive bars should be equal.Step V. Shade the bars of both thetypes with different shadings. Theirshadings are shown at the top rightcorner of the graph paper.

❏❏


WORKSHEET–29

1. (A) Nine times x = 9 × x = 9x

∴ The required form is ' 9x + 6 = 24.

2. (B) 5p

is the one-fifth of a number p.

So, the statement form of 5p

– 3 = 0 is

given by 'taking away 3 from one-fifthof a number p gives 0'.

3. (C) ∵ '3 times a' means 3a

∴ ' 3 times a is 39 means' 3a = 39.

4. (A) x + 5 = 6

or x = 6 – 5

(Transposing 5 to the right)

or x = 1.

5. (B) 7p +8 = 22

or 7p = 22 – 8


or 7p = 14

or p = 2 (Dividing both sides by 7).

6. (A) 5(n – 2) = – 4

or n – 2 = – 45

(Dividing both sides by 5)

or n = 2 – 45

(Transposing – 2 to the right)

or n = 10 4

5−

= 65 .

7. (B)52 y = 10

5Chapter

SIMPLE EQUATIONS

or y = 10 × 25

(Multiplying both sides by 25 )

or y = 4.

8. (D) x = 1

or x – 1 = 1 – 1

(Subtracting 1 from both the sides)

or x – 1 = 0.

9. (A) Let us take option (A).

6x + 6 = 0

LHS = 6x + 6

= 6( – 1) + 6

(Substituting x = –1)

= – 6 + 6 = 0

= RHS

Clearly, LHS = RHS, so x = –1 is assolution of 6x + 6 = 0.

10. (B) Let us substitute y = 6 in option (B)

3y

– 1 = 1 or 63

– 1 = 1

or 2 – 1 = 1

or 1 = 1 which is true.

So, y = 6 is as solution of 3y

– 1 = 1.

11. (B) Let us substitute x = – 3 in option (B)

4x + 3 = – 9

4 × (– 3) + 3 = – 9

or – 12 + 3 = – 9

or – 9 = – 9

which is true. So, x = – 3 is the solutionof 4x + 3 = – 9.

53

I

S I M P L E E Q U A T I O N S

12. (D) x – 6 means 'take away 6 from x'

∴ x – 6 = 3 means 'taking away 6 fromx gives 3'.

13. (D) x – 6 = 1or x = 1 + 6 ( Transposing – 6)

or x = 7.

14. (C)3p

= 5 or p = 15

(Multiplying both sides by 3)15. (C) – 8 = 6 + 8 (x – 2)

or – 8 – 6 = 8 (x – 2)

or x – 2 = – 148 = –

74

or x = 2 – 74

= 14

.

16. (A) On transposing a number, the signof the number changes from positiveto negative and vice-versa.

17. (D) Division by zero is not defined.18. (A) Let unknown number be x. Then

6x

– 3 = 4

or6x = 4 + 3 = 7

or x = 6 × 7 = 42.

19. (C) Let unknown number be x. Then

52 x – 7 = 18

or52 x = 18 + 7 = 25

or x = 25 × 25 =

505 = 10.

20. (C) 72l

= 27

or l = 27 ×

27


or l = 449 .

WORKSHEET – 30

1. (i) x + 7 = 20

LHS = x + 7

= 13 + 7 (Substituting x = 13)

= 20 = RHS

As LHS = RHS, x = 13 is thesolution of x + 7 = 20.

(ii) x – 5 = 15

LHS = x – 5

= 10 – 5 (Substituting x = 10)

= 5

∵ LHS ! RHS

∴ x = 10 is not the solution of

x – 5 = 15.

(iii) x – 9 = 23

LHS = x – 9

= 32 – 9 (Substituting x = 32)

= 23 = RHS

∵ LHS = RHS

∴ x = 32 is the solution of x – 9 = 23.

(iv) 15 – x = – 2

LHS = 15 – x

= 15 – 13

(Substituting x = 13)

= 2

∵ LHS ! RHS

∴ x = 13 is not the solution of15 x = – 2

(v) 5x = 25

LHS = 5x

= 5 × 0 (Substituting x = 0)

= 0

∵ LHS ! RHS

∴ x = 0 is not the solution of

5x = 25.


(vi) 4 m – 12 = 4LHS = 4 m – 12

= 4 × 4 – 12 (Substituting m = 4)= 16 – 12 = 4= RHS

∵ LHS = RHS∴ m = 4 is the solution of4 m – 12 = 4.

2. (i) The given equation is 2x + 3 = 17

Here, LHS = 2x + 3

and RHS = 17

Let us take values of x till the LHSbecomes equal to the RHS as givenin the following table.

x LHS RHSRelation between

LHS and RHS

1 2×1+3= 5 17 LHS ! RHS

2 2 × 2 + 3 = 7 17 LHS ! RHS

4 2 × 4 + 3 = 1 1 17 LHS ! RHS

6 2 × 6 + 3 = 1 5 17 LHS ! RHS

7 2 × 7 + 3 = 1 7 17 LHS ! RHS

Clearly, LHS = RHS for x = 7.

So, x = 7 is the solution of

2x + 3 = 17.

(ii) The given equation is 4 m – 12 = 4

Here, LHS = 4 m – 12

and RHS = 4

Let us take the values of m till theLHS becomes equal to the RHS asshown in the following table:

m LHS RHSRelation betweenLHS and RHS

1 4 × 1 – 1 2 = – 8 4 LHS ! RHS3 4 × 3 – 1 2 = 0 4 LHS ! RHS

4 4 × 4 – 1 2 = 4 4 LHS = RHS

Clearly, LHS = RHS for m = 4

So, m = 4 is the solution of

4 m – 12 = 4.

3. (i) Three times a number m = 3 × m= 3 m

So, 'three times a number m is 20'means

3 m = 20.

(ii) Sum of a and 8 = a + 8

So, ‘sum of a and 8 is 16’ means

a + 8 = 16.

4. (i) The statement form of ‘x + 7 = 9’ is‘sum of x and 7 is 9’.

(ii) The statement form of

‘4m – 7 = 18’ is

‘if you take away 7 from 4 times m,you get 18’.

(iii) The statement form of 4 ’‘ 103x = is

‘four-third of x is 10’.

(iv) The statement form of

‘3p – 1’ = 24’ is

‘if you take away 1 from 3 times p,you get 24’.

5. (i) x – 3 = 2Adding 3 to both the sides, we get

x – 3 + 3 = 2 + 3or x = 5.

(ii) x + 19 = 20Subtracting 19 from both sides, weget

x + 19 – 19 = 20 – 19or x = 1.

(iii) x – 3 = – 10Adding 3 to both sides, we get

x – 3 + 3 = – 10 + 3

or x = – 7.

55

I


(iv) x – 5 = 5

Adding 5 to both the sides, we get

x – 5 + 5 = 5 + 5

or x = 10.

6. (i) 5x = 25

or55x

= 255


or x = 5

(ii) 2x = 0

or22x

= 02


or x = 0

(iii) 8x = 72

or88x

= 728


or x = 9.

7. (i) 7x

= 5

or 7x

× 7 = 5 × 7

(Multiplying both the sides by 7)

or x = 35.

(ii) 10 = 4x

or 4x

= 10

or 4x

× 4 = 10 × 4

(Multiplying both the sides by 4)

or x = 40.

(iii) – 8x

= 8

or – 8x

× (– 8) = 8 × (– 8)

[Multiplying both the sides by (– 8)]

or x = – 64.

WORKSHEET – 311. (i) 6x + 6 = 30

or 6x = 30 – 6


or 6x = 24

or 66x

= 246


or x = 4.

(ii) 4 m – 4 = 24

or 4 m – 4 + 4 = 24 + 4

(Adding 4 to both sides)

or 4 m = 28

or4

4m

= 284


or m = 7

(iii) 5 m – 5 = 15

or 5 m – 5 + 5 = 15 + 5

(Adding 5 to both sides)

or 5 m = 20

or5

5m

= 205


or m = 4.

2. (i)49x

= 20

Multiplying both sides by 9, we get

49x

× 9 = 20 × 9

or 4x = 180

Dividing both sides by 4, we get

44x

= 180

4 or x = 45.


(ii)35x

= 3

Multiplying both sides by 53 , we

get

35x

× 53 = 3 ×

53

or x = 5.

(iii) 5x

= 715


5x

× 5 = 715 × 5

or x = 73 .

3. (i) 6 (x + 5) = 18


x + 5 = 186 = 3

Subtracting 5 from both the sides,we get

x = 3 – 5

or x = – 2.

(ii) 3 (x – 5) = – 21


x – 5 = 213

− = – 7

Adding 5 to both the sides, we get

x = – 7 + 5

or x = – 2.

(iii) 34 – 5 (y – 1) = 4

Transposing 34 to the right, we get

– 5 (y – 1) = 4 – 34 = – 30

Dividing both sides by – 5, we get

y – 1 = – 30– 5 =

305 = 6

Adding 1 to both sides, we get

y = 6 + 1

or y = 7.

4. Let the unknown number be a.

So, 15 + a = 45

Subtracting 15 from both sides, we get

a = 45 – 15

or a = 30.

5. Let the required three numbers be x,

x + 1 and x + 2. Then

x + x + 1 + x + 2 = 84

or 3x + 3 = 84

or 3x = 84 – 3 = 81

or x = 813 = 27

∴ x + 1 = 27 + 1 = 28

And x + 2 = 27 + 2 = 29

Hence the required numbers are 27, 28

and 29.

6. Let the number be x.

Then 57 – x = 10

or x = 57 – 10

= 47.

7. Let the angles be a, 2a and 3a.

Using the angle sum property of atriangle, we get

a + 2a + 3a = 180°

or 6a = 180°

or a = 30°

∴ 2a = 2 × 30° = 60°

And 3a = 3 × 30° = 90°

Hence, the angles are 30°, 60° and 90°.

57

I


WORKSHEET – 32

1. Let the one number be x

Then the other number = 9x

According to the question,

x + 9x = 200

or 10x = 200

or x = 20010 = 20

∴ 9x = 9 × 20 = 180

So the required numbers are 20 and180.

2. (i) x = 2 (Given)

Adding 5 to both sides,

x + 5 = 2 + 5

x + 5 = 7.

(ii) x = 2 (Given)

Multiply both sides by 4. 4x = 8.

Subtract 2 from both sides

4x – 2 = 8 – 2

4x – 2 = 6

Thus the two equations are

x + 5 = 7 and 4x – 2 = 6.

3. (i) 8 (2 – x) = 48

Divide both sides by 8,

2 – x = 488 = 6

Multiply both sides by (– 1),

x – 2 = – 6

Add 2 to both sides.

x = – 6 + 2

or x = – 4.

(ii) 80 – 5 (y – 1) = 0

Subtract 80 from both sides.

– 5 (y – 1) = – 80

Multiply both sides by ( )15

− .

y – 1 = 16


y = 17

So required solution is y = 17.

(iii) 28 = 4 + 3 (x + 5)


24 = 3 (x + 5)

Divide both sides by 3.

8 = x + 5


3 = x

So required solution is x = 3.

(iv) 0 = 16 + 2 (m – 3)

Subtract 16 from both sides

– 16 = 2 (m – 3)

Divide both sides by 2.

– 8 = m – 3


– 5 = m

So required solution is m = – 5.

4. (i) 5x

= 12

Multiply both sides by 5.

x = 52 .

(ii)23

− x = 12

Multiply both sides by – 32 .

23

− × ( )32

− x = 12 × ( )32

−

or x = 362

−

or x = – 18.


(iii)xy = z

Multiply both sides by y.

x = yz.

(iv)axb = c

Multiply both sides by ba

.

axb ×

ba = c ×

ba

or x = bca .

5. (i) Difference of a and 6 is 18i.e., a – 6 = 18.

(ii) The number m is 10 more than 19i.e., m = 10 + 19.

(iii) Nine times m plus 8 gives you 98i.e., 9m + 8 = 98.

(iv) One fourth of a number x equals 1

i.e., 4x

= 1.

6. Let the number be x.

Thrice x means 3 times of x i.e., 3xAccording to the question,

40 – 3x = – 50

or – 3x = – 50 – 40


or – 3x = – 90

or – 3– 3

x=

– 90– 3

(Dividing throughout by – 3)

or x = 903 = 30

Thus, the number is 30.

7. Let the number be y.

5 times of y = 5 × y = 5y

8 times of y = 8 × y = 8y


8y – 5y = 60

or 3y = 60

or 33y

= 603


or y = 20

Thus, the required number is 20.

WORKSHEET – 33

1. (i) 8x = 48


x = 488 or x = 6.

(ii) 7x = 35


x = 357

or x = 5.

(iii) 5x = 65


x = 655 or x = 13.

2. (i) 3m + 17 = 32

or 3m = 32 – 17 = 15


or m = 153

(Divding both sides by 3)or m = 5.

(ii) 11m + 9 = 42or 11m = 42 – 9 (Transposing 9 to the right)or 11m = 33

or m = 3311

(Dividing both sides by 11)or m = 3.

59

I


(iii) 7m + 192 = 13

or 7m = 13 – 192


= 26 19

2−

= 72

or m = 7

2 7×(Dividing both sides by 7)

or m = 12 .

3. (i)89x

= 32

⇒89x

× 98 = 32 ×

98


⇒ x = 4 × 9

⇒ x = 36.

(ii)65x

= 30

⇒65x

×56 = 30 ×

56


⇒ x = 5 × 5

⇒ x = 25.

(iii)1415

x=

730

⇒ 1415

x×

1514 =

730 ×

1514


⇒ x = 714 ×

1530 =

12 ×

12

⇒ x = 14 .

4. (i) 6 (x + 3) = 48

or x + 3 = 486


or x + 3 = 8

or x = 8 – 3

(Transposing 3 to RHS)

x = 5.

(ii) 3 (x – 8) = – 27

or x – 8 = 273

− = – 9


or x = – 9 + 8

(Transposing – 8 to RHS)

or x = – 1.

(iii) 38 – 6 (y – 1) = 8

or – 6 (y – 1) = 8 – 38 = – 30


or y – 1 = – 30– 6

(Dividing both sides by – 6)

or y – 1 = 5

or y = 5 + 1


or y = 6.5. Let Ritu's age be x years

As Geeta is 8 years older than Ritu,Geeta’s age = (x + 8) years

∵ Sum of their ages = 26 years∴ x + x + 8 = 26or 2x + 8 = 26or 2x = 26 – 8 = 18

(Transposing 8)or x = 9

(Dividing throughout by 2)


∴ x + 8 = 9 + 8 = 17

Hence, Ritu is of 9 years and Geeta isof 17 years.

6. The number of girls is 60 more thanthat of the boys.So, the number of girls is greater.Let the number of boys = xThen the number of girls = 60 + x

∴ Total number of students

= x + 60 + x

= 2x + 60

But the total number of students

= 1200 (Given)

∴ 2x + 60 = 1200

or 2x = 1200 – 60

= 1140


or x = 1140

2 = 570


∴ 60 + x = 60 + 570

= 630.

Hence, number of boys = 570

and number of girls = 630.

7. (i) m – 40 = 6 or m = 6 + 40.

(ii) x – 5 = 23 or x = 5 + 23.

(iii) 4x

= 7.

(iv) 3x

– 10 = 30.

8. (i) 5ax

= b

Multiplying both sides by 5a , we

get

5ax

× 5a = b ×

5a or x =

5ba .

(ii)axp = c

Multiplying both sides by pa , we

get

axp ×

pa = c ×

pa or x =

cpa .

(iii) 2xy = 3z

Multiplying both sides by 2y, weget

2xy × 2y = 3z × 2y

or x = 6yz.

(iv)29x

= b + 7

Multiplying both sides by 92 , we

get

29x

× 92 = (b + 7) ×

92

or x = ( 7) 9

2b + ×

.

WORKSHEET – 34

1. (i) Let each of the base angle be a.According to the angle sumproperty of a triangle,

a + a + 35° = 180°or 2a + 35° = 180°This is the required equation.Here, 2a = 180° – 35°

(Transposing 35° to RHS)or 2a = 145°

or a = 145

2

ο

(Dividing both sides by 2)or a = 72.5°So, each of the base angles is 72.5°.

61

I


(ii) Let the number be xTwice x = 2xThrice x = 3xAccording to the question,

2x + 3x = 50This is the required equation.Here, 5x = 50

or x = 505 = 10

(Dividing both sides by 5)So the number is 10.

(iii) Let the one number be x

Then the other number = 4x

Sum of these two numbers = 200

∴ x + 4x

= 200

This is the required equation.

Here,4

4x x+

= 200

or54x

= 200

or54x

× 45 = 200 ×

45

(Multiplying both sides by 45

)

or x = 40 × 4 = 160

∴ 4x

= 160

4 = 40

Thus the required numbers are 160and 40.

(iv) Let Ravi has ` xThen Reema will have ` 2x

Sum of their rupees = 150∴ x + 2x = 150This is the required equation.or 3x = 150

or x = 150

3 = 50

So Ravi has ` 50.

2. (i) 35 – 5(y – 4) = 5

Subtracting 35 from both sides, weget

– 5 (y – 4) = – 30

or 5 (y – 4) = 30


y – 4 = 305 = 6

or y = 6 + 4


or y = 10.

(ii) – 8 = 10 (p – 2)

or – 8

10 = p – 2


or 2 – 45

= p

(Transposing – 2 to LHS)

or p = 10 – 4

5

or p = 65 .

(iii) 40 = 4 + 3 (x + 5)

or 40 – 4 = 3(x + 5)

or x + 5 = 363 = 12


or x = 12 – 5


or x = 7.

(iv) 50 = 16 + 2 (m – 5)

or 50 – 16 = 2 (m – 5)

or m – 5 = 342 = 17

or m = 5 + 17

or m = 22.


3.(i) 4x – 3 = 13Here, LHS = 4x – 3

= 4 × 1 – 3(Substituting x = 1)

= 1 ! RHSSo x = 1 does not satisfy the equation4x – 3 = 13.

(ii) 5p + 2 = 17Here, LHS = 5p + 2

= 5 × 3 + 2 (Substituting p = 3)

= 17 = RHSSo p = 3 satisfies the equation5p + 2 = 17

(iii) 5x = 25Here LHS = 5x

= 5 × 5(Substituting x = 5)

= 25 = RHSSo x = 5 satisfies the equation 5x = 25.

4. Let Bulbul has n marbles.Then Kanika has 10 × n + 7 = 10n + 7marblesAccording to the question,

n + 10n + 7 = 29or 11n = 29 – 7 = 22


or n = 2211 = 2

∴ 10n + 7 = 10 × 2 + 7 = 20 + 7= 27.

Therefore, Bulbul has 2 marbles andKanika has 27 marbles.

5. Let breadth = xSo, length = 12 + xPerimeter of a rectangle

= 2 × (length + breadth)

= 2 × (12 + x + x)

= 2 × (12 + 2x)

= 2 × 12 + 2 × 2x

= 24 + 4x.

But it is given that the perimeter of thetriangle is 48 cm.

∴ 24 + 4x = 48

or 4x = 48 – 24 = 24

or x = 244 = 6

∴ 12 + x = 12 + 6 = 18

Thus, length = 18 cm

and breadth = 6 cm.

6. (i) y – 15 = – 30

Transposing – 15 to RHS;

y = – 30 + 15

Thus y = – 15.

(ii) y + 90 = – 60

Transposing 90 to RHS;

y = – 60 – 90

Thus y = – 150.

WORKSHEET – 35

1. (i) x – 3 = 40The statement form of this equationis ‘Take away 3 from x gives 40’.

(ii) 7p + 2 = 9The statement form of this equationis ‘ The sum of 7 times p and 2 is 9’.

(iii) 3m + 5 = 15The statement form of this equationis ‘The sum of 3 times m and 5 is15’.

(iv)42x

= 10

The statement form of this equationis ‘Half of four times x is 10’.

63

I


(v) 7x – 2 = 3

The statement form of this equationis ‘Take away 2 from seven times xgives 3’.

2. (i) y + 17 =

37

Here, LHS = y + 17

= 37 +

17

(Substituting y = 37 )

= 47

Clearly LHS ! RHS

So, y = 37 is not the solution of

y + 17 =

37 .

(ii) 7 – x = 4Here LHS = 7 – x

= 7 – 2 = 5 (Substituting x = 2)

Clearly LHS ! RHSSo x = 2 is not the solution of7 – x = 4.

(iii) 2p = 18Here, LHS = 2p

= 2 × 9 = 18 (Substituting p = 9)Clearly LHS = RHSSo p = 9 is the solution of 2p = 18.

(iv) 5x = 125Here LHS = 5x

= 5 × 9 = 45 (Substituting x = 9)Clearly LHS ! RHSSo x = 9 is not the solution of5x = 125.

3. (i) 9x + 5 = 4x + 30Transposing 5 to RHS and 4x to LHSsimultaneously, we get

9x – 4x = 30 – 5or 5x = 25or x = 5.

(ii) 5x + 2 = 3x + 12Transposing 2 to RHS and 3x to LHSsimultaneously, we get

5x – 3x = 12 – 2or 2x = 10or x = 5.

(iii) b + 3 = 33 – bTransposing 3 to RHS and – b toLHS simultaneously, we get

b + b = 33 – 3or 2b = 30or b = 15.

(iv) 4t + 5 = t + 15Transposing 5 to RHS and t to LHSsimultaneously, we get

4t – t = 15 – 5or 3t = 10

or t = 103

(v) 2 – (3 – x) = – 4or 2 – 3 + x = – 4or – 1 + x = – 4Transposing – 1 to RHS, we get

x = – 4 + 1or x = – 3

4. (i) 3 (x + 6) = 4 (2x – 8)or 3x + 18 = 8x – 32or 3x – 8x = – 32 – 18or – 5x = – 50

or x = – 50– 5

= 505

or x = 10.


(ii)53m

+ 7 = 4m

– 1

Multiplying both sides by LCM of 3and 4 = 12, we get

20m + 84 = 3m – 12

or 20m – 3m = – 12 – 84

or 17m = – 96

or m = – 9617 .

5. Ratio of ages of Ram and Shyam is

3 : 5, i.e., 35 . It means ‘Ram’s age is

equal to three-fifth of Shyam’s age’.

Let Shyam’s age (in years) = x

Then Ram’s age = 35 × x =

35 x

Since sum of their ages is 40 years

∴ x + 35

x = 40

or5 3

5x x+

= 40

or85x

= 40

or x = 40 × 58 = 25


∴35

x = 35 × 25 = 15.

Hence Ram’s age is 15 years andShyam’s age is 25 years.

6. (i) Let Meenu’ previous weight = x kg

So, after losing 15 kg, her weight= (x – 15) kg

Now, according to the givencondition,

x – 15 = 75


Here x = 75 + 15 = 90.


So Meenu’s previous weight was90 kg.

(ii) Let 8% of x be 30.

∴8

100 × x = 30


Multiplying both sides by 100

8, we

get

8100 × x ×

1008 = 30 ×

1008

or x = 3000

8or x = 375.

(iii) Let Priti had m mangoes originally.

After giving 18 mangoes to Shanu,Priti was left with (m – 18) mangoes.

But Priti was left with 45 mangoes.

∴ m – 18 = 45


Here m = 45 + 18 = 63


So, Priti had 63 mangoes originally.

WORKSHEET – 36

1. (i) y + 7 = 8Subtracting 7 from both sides, weget

y + 7 – 7 = 8 – 7or y = 1.

(ii) 16 = x + 6Subtracting 6 from both sides, weget

16 – 6 = x + 6 – 6or 10 = xor x = 10.

65

I


(iii) 14 = b + 2


14 – 2 = b + 2 – 2

or 12 = b

or b = 12.

(iv) y – 10 = 18


y – 10 + 10 = 18 + 10

or y = 28.

(v) 7x = 56


77x

= 567

or x = 8.

(vi) 5t = 30


55t

= 305

or t = 6.

(vii) 75 = 5y


755 =

55y

or 15 = y or y = 15.

2. (i) 2y

= 10


2y

× 2 = 10 × 2 or y = 20.

(ii) 4c

= 10


4c

× 4 = 10 × 4 or c = 40.

(iii) 3x – 5 = 10


3x – 5 + 5 = 10 + 5

or 3x = 15.


33x

= 153

or x = 5.

(iv) m – (– 4) = 9

or m + 4 = 9


m + 4 – 4 = 9 – 4

or m = 5.

(v) 2 – (3 – a) = – 4

or 2 – 3 + a = – 4

or a – 1 = – 4


a – 1 + 1 = – 4 + 1

or a = – 3.

3. (i) 2x – 3 = 9 – x

or 2x + x = 9 + 3 (Transposing)

or 3x = 12

or x = 123 = 4.

(ii) 5n – 9 = n + 7

or 5n – n = 9 + 7 (Transposing)

or 4n = 16

or n = 4.

(iii) 18 – 5d = 3d – 6

or – 5d – 3d = – 6 – 18

(Transposing)

or – 8d = – 24

or d = – 24– 8 =

248

or d = 3.


(iv) x + 4x + 5x = 14

or 1x + 4x + 5x = 14

or 10x = 14

or x = 1410 = 1.4.

(v) 2t + 3t + 5 = 20

or 5t + 5 = 20

or 5t = 20 – 5 = 15

or t = 155 = 3.

4.4.4.4.4. Let Roma’s present age be x years.After 15 years, her age = (x + 15) yearsAlso, after 15 years, her age

= 4 times x= 4x

Therefore, 4x = x + 15or 3x = 15 (Transposing x to LHS)

or x = 153 = 5.

Hence, Roma’s present age is 5 years.5.5.5.5.5. (i) Length of each part

= Length of ribbonNumber of parts

= 5x

m.

(ii) If the length of each part is 10 m,

Then 5x

= 10

or x = 5 × 10 = 50.So, the length of whole piece is50 m.

6.6.6.6.6. Let the two consecutive numbers be xand x + 1.∵ Sum of these = 139∴ x + x + 1 = 139or 2x + 1 = 139or 2x = 139 – 1 = 138 (Transposing 1 to RHS)

or22x

= 138

2

(Dividing both sides by 2)or x = 69∴ x + 1 = 69 + 1 = 70.Hence, the numbers are 69 and 70.

7.7.7.7.7. Let the number be p.∴ Five times p = 5pAccording to the given condition,

5p + 2 = 37or 5p = 37 – 2 = 35 (Transposing 2 to RHS)

or55p

= 355

(Dividing both sides by 5)or p = 7So, the required number is 7.

❑❑

67ENIL AS DN A GN L SE

WORKSHEET –37

1. (C) Let the complement of 53° be x.

Then x + 53° = 90°

∴ x = 90° – 53° = 37°.

2. (A) The sum of two complementaryangles is a right angle i.e., 90°.

3. (D) Let each of two complementaryangles be x.

Then x + x = 90°

or 2x = 90°

∴ x = 90

2

ο= 45°.

4. (B) As the sum of two complementaryangles is 90°, each of them will beacute.

5. (C) Let the required angle = 2x

Then the other angle = 22x

= x

So, 2x + x = 90° ⇒ 3x = 90°

⇒ x = 903

° = 30°

∴ 2x = 2 × 30° = 60°.

6. (A) A line is obtained when a linesegment is extended on both sides soa line has no end points.

7. (D) x + 62° = 90°

⇒ x = 90° – 62° = 28°.

8. (B) Let one of two complementary angles = x

Then the other one = 90° – x


x – (90° – x) = 18°

6Chapter

LINES AND ANGLES

or 2x = 18° + 90° = 108°

or x = 108

2°

= 54°

∴ 90° – x = 90° – 54° = 36°.9. (D) Sum of angles of a linear pair

= 180°.10. (A) Every line segment has two end

points.11. (A) Since ∠1 and ∠2 are vertically

opposite angles∴ ∠1 = ∠2.

12. (C)Sum of two supplementary angles = 180°Here, 40° + 140° = 180°.

13. (A) Supplement of 71° = 180° – 71°= 109°.

14. (A) Let one of two supplementaryangles = x

∴ Other one = 180° – xBut x – (180° – x) = 88° (Given)∴ 2x – 180° = 88°or 2x = 88° + 180° = 268°

or x = 268

2

ο = 134°

∴ 180° – x = 180° – 134° = 46°.15. (C) ∵ ∠1 + ∠2 = 180° ! 90°

∴ ∠1 and ∠2 do not form a pair ofcomplementary angles.

16. (A) Two intersecting lines pass througheither one or infinitely many commonpoints.

17. (B) Since ∠1 and ∠2 form a pair ofinterior angles on the same side oftransversal n. ∴ ∠1 + ∠2 = 180°.


18. (D) Interior angles are ∠3, ∠4, ∠5 and∠6. ∠4 and ∠6, ∠3 and ∠5 are thepairs of alternate interior angles.

WORKSHEET –38

1. (i) ∵ 60° + 30° = 90°So 60° and 30° form a pair of compl-ementary angles:

(ii) 70° + 30° = 100° ! 90°; 70° and 30°do not form a pair of compl-ementary angles.

(iii) 35° + 45° = 80° ! 90°; 35 and 45° donot form a pair of complementaryangles.

(iv) 60° + 20° = 80° ! 90°; 60° and 20° donot form a pair of complementaryangles.

2. Let the angle be x. Then its complementis 90° – x.But x = 90° – x (Given)∴ 2x = 90°or x = 45°.

3. We know that the sum of twocomplementary angles is 90°.

(i) Complement of 60° = 90° – 60° = 30°.

(ii) Complement of 33° = 90° – 33° = 57°.

(iii) Complement of 82° = 90° – 82° = 8°.

4. Let one angle be x. Then the other angleis x + 18°.

Now x and x + 18° are complementaryangles.

∴ x + x + 18° = 90°

⇒ 2x = 90° – 18° = 72°

⇒ x = 72

2

ο = 36°

∴ x + 18° = 36° + 18° = 54°

So the measures of the two angles are36° and 54°.

5. Yes. Since the sum of angles of a linearpair is 180°, therefore a linear pair isan example of supplementary angles.

6. We know that sum of the supplemen-tary angles is 180°.

(i) Supplement of 58° = 180° – 58°= 122°.

(ii) Supplement of 113° = 180° – 113°= 67°.

(iii) Supplement of 125° = 180° – 125°= 55°.

7. (i) Vertically opposite angles are:

∠POY and ∠QOX;

∠POX and ∠QOY

(ii) Vertically opposite angles are:

∠QON and ∠POM;

∠QOM and ∠PON.

8. (i) x and 140° form a linear pair ofangles with AB (see figure).

∴ x + 140° = 180°

⇒ x = 180° – 140°

⇒ x = 40°.

(ii) x and (x + 18°) form a linear pair ofangles with MN (see figure).

∴ x + x + 18° = 180°

⇒ 2x + 18° = 180°

⇒ 2x = 180° – 18°

⇒ 2x = 162°

⇒ x = 162

2°

⇒ x = 81°


9. (i) p + 110° = 180° (Linear pair)⇒ p = 180° – 110° = 70°

q = p = 70°(Corresponding angles)

x = q (Corresponding angles)= 70°.

(ii) x and 130° form a pair of correspon-ding angles.∴ x = 130°.

(iii) Angles n and x form a pair ofcorresponding angles (see figure).∴ n = xAngles 2x and nform a linear pair ofangles

∴ 2x + n = 180°

⇒ 2x + x = 180° (∵ n = x)

⇒ 3x = 180°

⇒ x = 180

3

ο ⇒ x = 60°.

(iv) x = y (pair of corresponding angles)46° + y = 180°

(Linear pair of angles)

⇒ 46° + x = 180° (∵ x = y)⇒ x = 180° – 46°

⇒ x = 134°.

10. There are two lines p and q; and l istheir transversal (see figure).

All exterior angles are:

∠1, ∠2, ∠7 and ∠8.

All interior angles are:

∠3, ∠4, ∠5 and ∠6.

WORKSHEET –39

1. We know that

(a)A pair of angles forms complemen-tary angles if their sum is 90°.

(b)A pair of angles forms supplemen-tary angles if their sum is 180°.

(i) 120° + 60° = 180°

This pair is of supplementaryangles.

(ii) 45° + 45° = 90°

This pair is of complementaryangles.

(iii) 110° + 70° = 180°


(iv) 36° + 54° = 90°


(v) 95° + 85° = 180°


(vi) 40° + 50° = 90°


2. We know that sum of an angle and itscomplement is 90°.

(i) Complement of 80° = 90° – 80°

= 10°.

(ii) Complement of 30° = 90° – 30°= 60°.

(iii) Complement of 12° = 90° – 12°= 78°.


3. We know that sum of an angle and itssupplement is 180°.(i) Supplement of 110° = 180° – 110°

= 70°.(ii) Supplement of 80° = 180° – 80°

= 100°.(iii) Supplement of 145° = 180° – 145°

= 35°. 4.(i)

∠1 and ∠2 are adjacent anglesbecause they have a common vertexO, a common arm OC and non-common arms OA and OB (seefigure) are on either side of thecommon arm OC.

(ii) ∠1 and ∠2 are adjacentangles because they havea common vertex O, acommon arm OC andnon-common arms OAand OB (see figure) are on eitherside of the common arm OC.

(iii) ∠1 and ∠2 are not adjacent anglesbecause they have no common arm.

(iv) ∠1 and ∠2 are not adjacent anglesbecause they have no common arm.

5. A pair of angles becomes a linear pairif it follows the conditions given below:(a) The angles have a common vertex;(b) The angles have a common arm;(c) The non-common arms are on

opposite sides of the common arm;and

(d) The non-common arms are oppositerays.(i) Two acute angles cannot make

a linear pair(ii) Two right angles can make a

linear pair.

6.(i) y = 118° (vertically opposite angles)∵ p y q and l is transversal

∴ z = y = 118° (Corresponding angles)x = z = 118° (Corresponding angles)

(ii) y + 63° = 180° (Linear pair of angles)⇒ y = 180° – 63° = 117°

∵ p y q and l is transversal

∴ x = y (Corresponding angles)

= 117°.

7. Pairs of the vertically opposite anglesare:

∠1 and ∠4; ∠2 and ∠3

∠5 and ∠7; ∠6 and ∠8

8. (i) Yes. In the given figure, a pair ofangles is formed with two equalcorresponding angles. So l y m.

(ii) Yes. In the given figure, alternateinterior angles are equal. So, l y m.

9. p y q and l is transversal

⇒ x + y = 180° (Linear pair of angles)

⇒ y = 180° – x


Also, 5x + 6° = y

(Pair of corresponding angles)

⇒ 5x + 6° = 180° – x (∵ y = 180° – x)

⇒ 6x = 180° – 6° = 174°

⇒ x = 174

6

ο

⇒ x = 29°.

WORKSHEET –401. (i) All pairs of alternate angles are:

∠1 and ∠6, ∠4 and ∠7, ∠2 and ∠5,∠3 and ∠8.

(ii) All pairs of corresponding anglesare:∠1 and ∠5, ∠2 and ∠6, ∠3 and ∠7,∠4 and ∠8

(iii) ∠3 and ∠5(iv) ∠4 and ∠6.

2. EF y GH and AB is transversal∴ x = 80° (Corresponding angles)AB y CD and GH is transversal∴ y = 80°

(Alternate interior angles).3. ∠EFD = ∠CFQ = 50°

(Verticallyopposite angles) ∠QFD + ∠CFQ = 180°

(Linear pair of angles)⇒ ∠QFD = 180° – 50° = 130° ∠CFE = ∠QFD = 130°

(Vertically opposite angles) ∠AEF = ∠CFQ = 50°

(Corresponding angles) ∠PEB = ∠AEF = 50°

(Vertically opposite angles)∠AEP = ∠CFE = 130°

(Corresponding angles) ∠FEB = ∠AEP = 130°

(Vertically opposite angles).

4. ∵ AB y CD and BE is transversal∴ x = ∠ECD (Corresponding angles)

= 30°∵ AB y CD and AC is transversal∴ y = ∠DCA (Alternate interior angles)

= 60°Since z and ∠ACE form a linear pair∴ z = ∠ACE = ∠ACD + ∠DCE

= 60° + 30°= 90°.

5. (i) Pairs of vertically opposite anglesare:∠1 and ∠3, ∠2 and ∠4.

(ii) Pairs of adjacent angles are:∠1 and ∠4, ∠4 and ∠3, ∠3 and ∠2,∠2 and ∠1.

6. In the given figure, AB and CD aretwo lines and EF is transversal.Since ∠CQP and ∠APE are correspon-ding angles of same measure each of120°.Therefore, AB y CDFurther, AB y CD and GH is transversal∴ ∠SRD = ∠PSR

(Alternate interior angles)or x = 105°.

7. (i) In the adjoining figure, AB y CD andPQ is transversal.

∴ ∠MND = ∠AMN(Alternate interior angles)

or x = 105°.


(ii) x = 130° (Corresponding angles)

(iii) Angles x and 50° which are shownin the given figure are correspond-ing angles.

∴ x = 50°.

(iv) Angles x and 110° which are shownin the given figure are correspo-nding angles

∴ x = 110°.8. (i)

Line n intersects lines l and m indistinct points. The two angles eachof measure 60° (see figure) arealternate interior angles.So l y m.

(ii) ∠QRD = ∠CRS = 110° (Vertically opposite angles)Clearly, ∠QRDand ∠PQB arec o r r e s p o n d i n gangles which areequal of measure110° each.∴ AB y CD.

WORKSHEET – 41

1. (i) The shown angles x and 63° in thegiven figure are vertically oppositeangles.∴ x = 63°.

(ii) The shown angles x and 95° in thegiven figure form a linear pair∴ x + 95° = 180°∴ x = 180° – 95° = 85°.

(iii) The shown angles x and 153° in thegiven figure form a linear pair.

∴ x + 153° = 180°∴ x = 180° – 153° = 27°.

(iv) l y m and n is transversal. Angles yand 80° are alternate interior angles∴ y = 80°

x + y = 180° (Linear pair of angles)

⇒ x + 80° = 180°

⇒ x = 180° – 80°

= 100°.2. (i) Let the complement of 13° be x1,

thenx1 + 13° = 90°

⇒ x1 = 90° – 13° = 77°.(ii) Let the complement of 78° be x2,

thenx2 + 78° = 90°

⇒ x2 = 90° – 78° = 12°.(iii) Let the complement of 35° be x3,

thenx3 + 35° = 90°

⇒ x3 = 90° – 35° = 55°.(iv) Let the complement of 18° be x4,

thenx4 + 18° = 90°

⇒ x4 = 90° – 18° = 72°.3. (i) Let the supplement of 152° be y1,

theny1 + 152° = 180°

⇒ y1 = 180° – 152° = 28°.(ii) Let the supplement of 105° be y2,

theny2 + 105° = 180°

⇒ y2 = 180° – 105° = 75°.


(iii) Let the supplement of 76° be y3, then

y3 + 76° = 180°

⇒ y3 = 180° – 76° = 104°.

(iv) Let the supplement of 128° be y4,then

y4 + 128° = 180°

⇒ y4 = 180° – 128° = 52°.

4. (i) Pairs of adjacent angles are:

∠1 and ∠2, ∠2 and ∠3, (∠1 + ∠2)and ∠3, ∠1 and (∠2 + ∠3).

There is no linear pair

(ii) Pairs of adjacent angles are:

∠1 and ∠2, ∠2 and ∠3, ∠3 and ∠4,∠4 and ∠1.

The linear pairs are:

∠1 and ∠2, ∠2 and ∠3, ∠3 and ∠4,∠4 and ∠1.

(iii) Pairs of adjacent angles are:

∠1 and ∠2, ∠2 and ∠3, ∠3 and ∠4 ,∠4 and ∠1.

Linear pairs are:

∠1 and ∠2, ∠2 and ∠3, ∠3 and ∠4,∠4 and ∠1.

(iv) Pairs of adjacent angles are:

∠1 and ∠2, ∠2 and ∠3, ∠3 and ∠4,∠4 and ∠5, ∠5 and ∠6, ∠6 and ∠1,(∠1 + ∠2) and ∠3, (∠2 + ∠3) and∠4, (∠3 + ∠4) and ∠5, (∠4 + ∠5)and ∠6, (∠5 + ∠6) and ∠1, (∠6 +∠1) and ∠2, (∠1 + ∠2) and ∠6, (∠2+ ∠3) and ∠1, (∠3 + ∠4) and ∠2,(∠4 + ∠5) and ∠3, (∠5 + ∠6) and∠4, (∠6 + ∠1) and ∠5.

Linear pairs are:

(∠1 + ∠2) and ∠3, (∠2 + ∠3) and ∠4,(∠3 + ∠4) and ∠5, (∠4 + ∠5) and ∠6,(∠5 + ∠6) and ∠1, (∠1,+ ∠6) and ∠2,

(∠1 + ∠2) and ∠6, (∠2 + ∠3) and ∠1,(∠3 + ∠4) and ∠2, (∠4 + ∠5) and ∠3,(∠5 + ∠6) and ∠4, (∠6 + ∠1) and ∠5.

5. (i) Since angles x and 112° form a linearpair∴ x + 112° = 180°⇒ x = 180° – 112° = 68°.

(ii) Since angles x and x – 100° form alinear pair∴ x + x – 100° = 180°⇒ 2x = 180° + 100° = 280°

⇒ x = 280

2

ο = 140°.

(iii) Since angles x and 123° are verticallyopposite angles∴ x = 123°.

(iv) Since angles x and (3x + 60°) form alinear pair∴ x + 3x + 60° = 180°⇒ 4x = 180° – 60° = 120°

⇒ x = 120

4

ο = 30º.

6. (i) ∵ l y m and n is transversal

∴ y = x

(Vertically opposite angles)

And 2x + y = 180°

(Interior angles on the same side of the transversal n)

⇒ 2x = 180° – y = 180° – x (∵ y = x)

⇒ 3x = 180°

⇒ x = 180

3

ο = 60°.


(ii) x + (x + 50°) = 180°(Interior angles on the same side of transversal)⇒ 2x = 180° – 50° = 130°

⇒ x = 130

2

ο = 65°.

(iii) ∵ l y m and n is transversal∴ y + 120° = 180°⇒ y = 180° – 120° = 60°Also, x = y = 60°.

(iv) Angles x and 60° are interior angleson the same side of transversal.∴ x + 60° = 180°⇒ x = 180° – 60° = 120°.

(v) Angles x and 25° are alternateinterior angles∴ x = 25°.

WORKSHEET –42

1. (i) Angles x, x + 20° and 60° are on thesame side of a straight line∴ x + x + 20° + 60° = 180°⇒ 2x + 80° = 180°⇒ 2x = 180° – 80°

= 100°

∴ x = 100

2

ο = 50°.

(ii) Angles x, 20°, 3x

and 160° form a

complete angle.

∴ x + 20° + 3x

+ 160° = 360°

⇒ x + 3x

= 360° – 160° – 20°

⇒43x

= 180°

⇒ x = 34

× 180°

= 3 × 45°⇒ x = 135°.

(iii) Angles x, 3x

and 120° form a

complete angle.

∴ x + 3x + 120° = 360°

⇒ x + 3x

= 360° – 120°

⇒ 43x

= 240°

⇒ x = 240° × 34

= 60° × 3⇒ x = 180°.

(iv) Angles x, (x + 20°) and (x + 10°) areon the same side of a straight lines

∴ x + x + 20° + x + 10° = 180°

⇒ 3x + 30° = 180°

⇒ 3x= 180° – 30° = 150°

⇒ x = 150

3°

= 50°.

2. (i) A line intersects the lines l and m,and a pair of corresponding anglesis equal.

So, l y m.

(ii) The upper horizontal line intersectsthe lines l and m, and a pair ofcorresponding angles is equal.

So l y m.

(iii) An oblique transversal intersects thelines l and m, and a pair of alternateinterior angles is equal.

So, l y m.(iv) 110° + 80° = 190° ! 180°


A transversal intersects the lines land m, and a pair of interior angleson the same side of the transversalis not supplementarySo l in not parallel to m.

3. (i) Angles b and 60° are verticallyopposite

∴ b = 60°

Angles a and b form a linear form

∴ a + b = 180°

⇒ a = 180° – b = 180° – 60°

(∵ b = 60°)

⇒ a = 120°

Angles a and c are verticallyopposite angles

∴ c = a = 120° (∵ a = 120°)

Thus a = 120°, b = 60°, c = 120°.

(ii) Angles b and 40° are verticallyopposite

∴ b = 40°

Angles 40°, c and 45° are on thesame side of a straight line.

∴ 40° + c + 45° = 180°

⇒ c = 180° – 40° – 45° = 95°

Angles a and (c + 45°) are verticallyopposite.

∴ a = c + 45°

= 95° + 45° (∵ c = 95°)

= 140°

Thus a = 140°, b = 40°, c = 95°.

4. (i) Angles x and (x + 28°) form alinear pair

∴ x + x + 28° = 180°

⇒ 2x = 180° – 28° = 152°

⇒ x = 152

2

ο = 76°.

(ii) Angles x, x, 3x and 3x form acomplete angle.∴ x + x + 3x + 3x = 360°⇒ 8x = 360°

⇒ x = 360

8

ο = 45°.

(iii) Angles x and (3x + 60°) form a linearpair.∴ x + 3x + 60° = 180°⇒ 4x = 180° – 60° = 120°

⇒ x = 120

4

ο = 30°.

(iv) y = x (Corresponding angles)Angles y and (5x + 6°) form a linearpair.

∴ y + 5x + 6° = 180°⇒ x + 5x + 6° = 180° (∵ y = x)⇒ 6x = 180° – 6° = 174°

⇒ x = 174

6

ο = 29°.

5. (i) 67° + 23° = 90°

⇒ The pair of angles 67° and 23° isof complementary angles.

(ii) 40° + 50° = 90°

⇒ The pair of angles 40° and 50° isof complementary angles.

(iii) 127° + 53° = 180°

⇒ The pair of angles 127° and 53° isof supplementary angles.

(iv) 113° + 67° = 180°

⇒ The pair of angles 113° and 67° isof supplementary angles.

6. Interior angles are: ∠3, ∠4, ∠5 and ∠6.

7. No, two obtuse angles cannot form alinear form.


WORKSHEET –431. AB y CD and an oblique line is

transversal (see figure).

∴ x = 60° (Corresponding angles)

CD y EF and an oblique line istransversal (see figure).

∴ y = 60° (Alternate interior angles)

2. (i) AB y CD and AD is transversal

∴ x = 110° (Corresponding angles)

AD y BC and DC is transversal

∴ y = 110° (Corresponding angles)

Thus, x = y = 110°.

(ii) PS y QR and SR is transversal

∴ a + 65° = 180°

(Interior angles on the same side of transversal SR)

⇒ a = 180° – 65° = 115°

SR y PQ and SP is transversal

∴ a + b = 180°

(Interior angles on the same side of transversal SP)

⇒ 115° + b = 180° (∵ a = 115°)

⇒ b = 180° – 115° = 65°

SR y PQ and RQ is transversal

∴ c + 65° = 180°(Interior angles on the same side of transversal RQ)

⇒ c = 180° – 65° = 115°.Thus, a = c = 115°, b = 65°.

3. (i) AC y BD and AB is transversal.Angles x and 3x are on the sameside of the transversal AB.∴ x + 3x = 180°⇒ 4x = 180°

⇒ x = 180

4

ο

⇒ x = 45°.(ii) Angle 3x is a right angle

i.e., 3x = 90°

∴ x = 903

ο = 30°.

(iii) Angles a and 80° are verticallyopposite angles∴ a = 80°.Angles x and a are correspondingangles.∴ x = a = 80°

Angles c and x form a linear pair.∴ c + x = 180°⇒ c = 180° – x = 180° – 80°

(∵ x = 80°)= 100°

Angles b and c are correspondingangles.∴ b = c = 100° (∵ c = 100°)Thus, a = 80°, b = 100°, c = 100°.

(iv) Angles a and 118° are correspondingangles.∴ a = 118°Angles c and 118° are correspondingangles.


∴ c = 118°

Angles b and c are alternate interiorangles.

∴ b = c = 118° (∵ c = 118°)

Thus, a = 118°, b = 118°, c = 118°.

4. ∠BCA, ∠ACD and ∠DCE are on thesame side of line BCE.

∴ ∠BCA + ∠ACD + ∠DCE = 180°

⇒ 74° + ∠ACD + 59° = 180°

⇒ ∠ACD + 133° = 180°

⇒ ∠ACD = 180° – 133°= 47°

Also, ∠BAC = 47°

(See figure)

In the given figure, line AC intersectsAB and CD. And so ∠BAC and ∠ACDare alternate interior angles each ofmeasure 47°.

Therefore, AB y CD.

5. (i) Angles y and 2x are verticallyopposite angles.

∴ y = 2x

l y m and n in transversal, so anglesy and 3x are on the same side of thetransversal.

∴ y + 3x = 180°

⇒ 2x + 3x = 180°

⇒ 5x = 180°

⇒ x = 180

5°

= 36°.

(ii) Angles x and (x + 30°) are on thesame side of the transversal.∴ x + x + 30° = 180°⇒ 2x = 180° – 30° = 150°

⇒ x = 150

2°

⇒ x = 75°.(iii) Angles 4x and 5x are on the same

side of the transversal.∴ 4x + 5x = 180°⇒ 9x = 180°

⇒ x = 180

9°

⇒ x = 20°.6. Angles a and 55° vertically opposite

angles

∴ a = 55°l y m and q is transversal.∴ x = a (Alternate interior angles)

= 55°p y q and m is transversal∴ b = x (Corresponding angles)

= 55°Thus, a = 55° and b = 55°.

❏❏


WORKSHEET –44

1. (C) In the adjoiningfigure, the sideopposite to thevertex B is CA.

2. (C) One median passes through eachvertex of a triangle. So number ofmedians is 3.

3. (A) We know that exterior angle of atriangle is equal to the sum of twoopposite interior angles.

∴ ∠ACD = ∠BAC + ∠ABC

= ∠1 + ∠2.

4. (D) Sum of angles of a triangle= 2 right angles

= 2 × 90° = 180°.

5. (C) 60° + 70° + ∠x = 180°

(Angle sum property)

⇒ ∠x = 180° – 130° = 50°.

6. (A) Exterior angle = Sum of two interior opposite angles

or 112° = 55° + x

∴ x = 112° – 55°

= 57°.

7. (A) ∵ Exterior angle = Sum of twocorresponding interior angles

∴ x = 55° + 50° = 105°.

8. (C) x + x + x = 180°


⇒ 3x = 180° ⇒ x = 180

3

ο= 60°.

7Chapter

TRIANGLES

9. (D) x = 70°

(Vertically opposite angles)

x + y + 60° = 180°


∴ y = 180° – 60° – 70° = 50°.

10. (C) ∠1 + ∠2 + ∠3 = y + y + y

⇒ 180° = 3y ⇒ y = 60°

11. (A) ∵ AB = AC

∴ ∠C = ∠B = x

(Angles opposite to

equal sides)

Further ∠A + ∠B + ∠C

= 180°


⇒ 130° + x + x = 180°

⇒ x = 180 130

2

ο ο−

= 25°.

12.(A) x + y + ∠B = 180°


⇒ x + y = 180° – 90° = 90°

(∵ ∠B = 90°).

13. (B) Let the given exterior angles is∠ACD.

79AIRT GN EL S

∴ ∠ACD = 90°

Further ∠BCA + ∠ACD = 180°


∴ ∠BCA = 180° – 90° = 90°

So Δ ABC is a right-angled triangle.

14. (C) We know that “The sum of anytwo sides of a triangle is greater thanthe third side.”

∴ In ΔABC, AB + BC > CA

15. (D) 4 cm + 3 cm > 6 cm

⇒ Sum of two sides > Third side.

16. (B) Let the third side be x.

Sum of two sides > Third side

⇒ 7 + x > 11

⇒ x > 4

⇒ x cannot be less than or equal to4 cm

⇒ x cannot be 3 cm.

17. (A) ∠P + ∠Q + ∠R = 180°

⇒ ∠R = 180° – 35° – 55° = 90°

So ΔPQR is a right

angled triangle

So, RP2 + QR2 = PQ2

(Pythagoras property).

18. (C) PR2 = PQ2 + QR2

(Pythagoras property)

⇒ QR2 = 152 – 92

= 225 – 81

= 144 = 12 × 12

⇒ QR = 12 cm.

WORKSHEET –45

1. There are three sides in the triangleABC.

These are AB, BC and CA.

There are three vertices in the triangleABC. These are A, B and C.

There are three interior angles in thetriangle ABC. These are ∠A, ∠B and∠C.

2. A line segment joining the mid-pointof a side of a triangle to its oppositevertex is called a median of triangle. Atriangle has three medians

3. We know that an exterior angle of atriangle is equal to the sum of interioropposite angles.

(i) 105° = 30° + x

∴ x = 105° – 30° = 75°.

(ii) 120° = x + 40°

∴ x = 120° – 40° = 80°.

4. A triangle is formed only when thetotal measure of the three angles is180°.

(i) Total measure of angles

= ∠A + ∠B + ∠C

= 30° + 60° + 90°

= 180°

So, the given measures form atriangle.

(ii) Total measure of angles

= ∠A + ∠B + ∠C

= 100° + 70° + 30°

= 200°

So, the given measures do not forma triangle.


5. (i) Let the missing angle be x.

Then 40° + 40° + x = 180°


or 80° + x = 180°

∴ x = 180° – 80° = 100°.

(ii) Let the missing angle be y.

Then 45° + 90° + y = 180°


∴ y = 180° – 45° – 90° = 45°.

6. (i) Yes, the triangle is possible becausethe sum of two sides (5 cm + 12 cm= 17 cm) is greater than the thirdside (13 cm).

(ii) Yes, the triangle is possible becausethe sum of two sides (3 cm + 6 cm =9 cm) is greater than the third side(7 cm).

7. (i) Sum of two sides

= BC + CA

= 10 cm + 10 cm = 20 cm

∴ BC + CA > AB

So, the given measures are the sidesof a triangle.

(ii) Sum of two sides

= AB + BC

= 8 cm + 8 cm = 16 cm

∴ AB + BC > CA

So, the given measures are the sidesof a triangle.

8. Let the other leg be x.

According to the Pythagoras propertyof a triangle,

x2 + 122 = 132

∴ x2 = 132 – 122 = 169 – 144

= 25

or x2 = 52

∴ x = 5 m.

9. x is an exterior angle of ΔBCD

∴ x = 100° + 25° = 125°

y is an exterior angle of ΔABE

∴ y = 20° + x = 20° + 125°

(∵ x = 125°)

= 145°.

10. x is the length of one leg of the givenright triangle.

∴ x2 + 32 = 52


or x2 = 5 × 5 – 3 × 3

or x2 = 25 – 9

or x2 = 16 = 4 × 4

∴ x = 4 cm.

WORKSHEET –46

1. The perpendicular line segment froma vertex of a triangle to its oppositeside is called an altitude of a triangle.A triangle has 3 altitudes.

The three altitudes do not always meetin the interior of the triangle.

2. (i) x + 55° + 45° = 180°


or x + 100° = 180°

∴ x = 180° – 100° = 80°.

(ii) x + 2x + 30° = 180°


or 3x + 30° = 180°

∴ 3x = 180° – 30° = 150°

81AIRT GN EL S

∴ x = 150

3

ο = 50°

And 2x = 2 × 50° = 100°.

3. (i) ∵ Sum of interior angles = 180°

∴ 45° + 90° + x = 180°

∴ x = 180° – 45°– 90° = 45°

Now,

y = x + 90°

(Exterior angle

property)

= 45° + 90° = 135°.

(ii) According to angle sum property ofa triangle,

50° + x + 20° + x = 180°

or 2x + 70° = 180°

∴ 2x = 180° – 70°

= 110°

∴ x = 110

2

ο

x = 55°

Now, y = 50° + x

(Exterior angle property)

= 50° + 55° (∵ x = 55°)

= 105°.

4. (i) An exterior angle of a triangle is thesum of interior opposite angles. Itis called ‘exterior angle property’.

110° = x + 30°


∴ x = 110° – 30° = 80°.

(ii) AD y BC and AC is transversal

∴ y = 70° (Alternate interior angles)

x + y + 80° = 180°


or x + 70° + 80° = 180°

(∵ y = 70°)

or x + 150° = 180°

∴ x = 180° – 150°

= 30°

5. (i) We know that according to the anglesum property of a triangle, “Sum ofinterior angles of a triangle is 180°”.

Here ∠P + ∠Q + ∠R

= 65° + 35° + 70°

= 170°

Which is not 180°, so the givenangles do not form a triangle.

(ii) We know that “To form a triangle,the sum of any two sides must begreater than the third side.”

Here YZ + XZ = 25 cm + 25 cm

= 50 cm

∴ YZ + XZ = XY

Clearly, YZ + XZ is not greater

than XY .

So, the given lengths of sides donot form a triangle.

6. In ΔABC, ∠A = 90°

So, BC is hypotenuse

∴ BC2 = AC2 + AB2



= 32 + 42 = 3 × 3 + 4 × 4

= 9 + 16 = 25

= 5 × 5

∴ BC = 5 cm.

7. Let the given isosceles triangle be ABCsuch that

AB = AC,

BC2 = 72 sq. m

And ∠A = 90°

According to the Pythagoras property,

BC2 = AB2 + AC2

or 72 = AB2 + AB2 (∵ AB = AC)

or 2 AB2 = 72

or AB2 = 722

= 36

= 6 × 6

∴ AB = 6 m

∴ AB = AC = 6 m.

WORKSHEET –47

1. (i) Sum of interior angles of a triangle = 180°


∴ 60° + 75° + x = 180°

or 135° + x = 180°

∴ x = 180° – 135°

or x = 45°.

(ii) An exterior angle of a triangle

= Sum of interior opposite angles


∴ 110° = 50° + x

or 50° + x = 110°

∴ x = 110° – 50°

or x = 60°.

2. (i) Sum of angles = 68° + 49° + 63°

= 180°

We know that total measure of threeangles of a triangle is 180°.

So, given angles form a triangle.

(ii) Sum of angles = 47° + 72° + 64°

= 183°

We know that total measure of threeangles of a triangle is 180°.

So, given angles do not form atriangle.

3. Let the required angle be x and theother two angles be y and z.

∴ x = y + z

Also x + y + z = 180°


or x + x = 180° (∵ x = y + z)

or 2x = 180°

∴ x = 180

2

ο

or x = 90°.

4. Let each of the two equal angles be xand the third one be y. Then

x = 2y

And x + x + y = 180°


or 2y + 2y + y = 180°

or y = 180

5

ο = 36°

∴ x = 2y = 2 × 36° = 72°

Hence, all the angles are 72°, 72° and36°.

5. Angles are in the ratio 3 : 4 : 1

Let the angles be 3x, 4x and x

83AIRT GN EL S

∴ 3x + 4x + x = 180°


∴ x = 180

8

ο = 22.5°

∴ 3x = 3 × 22.5° = 67.5°

And 4x = 4 × 22.5° = 90°

Hence, the angles are 67.5°, 90° and22.5°.

6. (i) Let the third angle be x.

30° + 80° + x = 180°


or 110° + x = 180°

∴ x = 180° – 110° = 70°.(ii)Let the third angle be y.

40° + 40° + y = 180°

or 80° + y = 180°

∴ y = 180° – 80° = 100°.

7. Let the required angle be x.

We know that one angle of a righttriangle is of 90°.

Now, x + 90° + 72° = 180°

or x + 162° = 180°

∴ x = 180° – 162°

or x = 18°.

8. Let each of three equal angles be x.

Then x + x + x = 180°

(Angle sum property of a triangle)

or 3x = 180°

∴ x = 180

3

ο = 60°.

So, all the angles are 60°, 60° and 60°.

9. No, it is not possible because in thiscase the third angle is of 0° but measureof each of the angles of a triangle mustbe a positive quantity.

10. The two angles are in the ratio 2 : 8

Let these angles be 2x and 8x.

Now 2x + 8x + 80° = 180°


or 10x = 180° – 80° = 100°

∴ x = 100

10

ο = 10°

∴ 2x = 2 × 10° = 20°

And 8x = 8 × 10° = 80°

Thus, the measures of the angles are80°, 20° and 80°.

11. No, because in this case sum ofmeasures of the three angles is morethan 180°.

WORKSHEET –48

1. Measure of each angle of an equilateraltriangle is 60°.

2. (i) Two sides of the given triangle areequal.

∴ x = 40°

(Angles opposite to equal sides)

(ii) In the given figure AB = AC

∴ ∠C = ∠B = x


Further ∠DAC = ∠B + ∠C


or 115° = x + x = 2x (∵∠C = x )

∴ x = 115

2

ο = 57.5°.


3. Let each of the required angles be x.

We know that an exterior angle of atriangle is equal to the sum of interioropposite angles.

∴ 100 = x + x

or 2x = 100°

∴ x = 50°.

4. Let ∠A = x (see figure)

Then, ∠B = 4x

Now, ∠ACD = ∠A + ∠B


or 110° = x + 4x = 5x

∴ x = 110

5

ο = 22°

∴ 4x = 4 × 22° = 88°

Further ∠BCA + ∠ACD = 180°

(Linear pair)

or ∠BCA + 110° = 180°

∴ ∠BCA = 180° – 110°

= 70°

Thus, the required angles are 22°, 88°and 70°.

5. (i) Sum of the given angles

= 70° + 80° + 30° = 180°.

We know that ‘The total measure ofangles of a triangle is 180°.’

So, the given measures can be thethree angles of a triangle.

(ii) Sum of the given angles

= 36° + 48° + 80° = 164°

We know that ’The total measure ofangles of a triangle is 180°.’

So, the given measures cannot bethe angles of a triangle.

6. Yes, each angle may be 60°.

7. Let the measure of the third angle be x

So 80° + 10° + x = 180°


or 90° + x = 180°

∴ x = 180° – 90° = 90°.

As one angle of the triangle is 90°, thetriangle is right-angled triangle.

8. (i) ∠A and ∠C are angles opposite tothe equal sides.

∴ ∠A = ∠C = x

Now ∠A + ∠B + ∠C = 180°


or x + 90° + x = 180°

∴ x = 902

ο= 45°.

(ii) Sides AC and AB are opposite tothe equal angles of given triangle.

∴ AC = AB

or x = 4.

9. Let the two angles which are in theratio 3 : 8 be 3x and 8x respectively.

Now 70° + 3x + 8x = 180°


∴ 11x = 180° – 70°

= 110°

∴ x = 11011

° = 10°

∴ 3x = 3 × 10° = 30°

and 8x = 8 × 10° = 80°

Thus, the measures of the other twoangles of the triangle are 30° and 80°.

10. Let the measures of the triangle be 2x,3x, and 5x.

Now 2x + 3x + 5x = 180°


or 10x = 180°

85AIRT GN EL S

∴ x = 180

10

ο = 18°

∴ 2x = 2 × 18° = 36°; 3x = 3 × 18° = 54°;and 5x = 5 × 18° = 90°.

Thus, the required angles are 36°, 54°and 90°.

11. x + 80° + 50° = 180°


or x + 130° = 180°

∴ x = 180° – 130° = 50°.

WORKSHEET –49

1. No. In this case, angle sum propertyof a triangle do not hold.

2. ∵ AB = AC

∴ ∠ACB = ∠ABC

But ∠ACB + 100° = 180°

∴ ∠ACB = 180° – 100° = 80°

Now ∠ACB + ∠ABC + x = 180°


or 80° + 80° + x = 180°

(∵ ∠ACB = ∠ABC)

∴ x = 180° – 160° = 20°.

3. In the given figure, 70° is an exteriorangle and interior angles opposite to itare 3x and 4x.

∴ 70° = 3x + 4x

(Exteriror angle property)

or 7x = 70°

∴ x = 707

° = 10°.

4. Yes. In every triangle at least two ofthe angles are acute.

5. (i) ∠A + ∠B + ∠C = 180°

(Sum of angles of a triangle is 180°)

⇒ 70° + 30° + ∠C = 180°

⇒ ∠C = 180° – 100°= 80°.

(ii) ∠A + ∠B + ∠C = 180°


⇒ 120° + 20° + ∠C = 180°

⇒ ∠C = 180° – 140°= 40°.

6. (i) Let in a triangle ABC;

AB = 9, BC = 60 and CA = 61

Here CA2 = 612 = 61 × 61= 3721

And AB2 + BC2 = 92 + 602

= 9 × 9 + 60 × 60

= 81 + 3600 = 3681

Clearly, CA2 ! AB2 + BC2

So, the given measures will not forma triplet.

(ii) Let in a triangle PQR;

PQ = 7, QR = 10 and RP = 6.

Here QR2 = 102 = 10 × 10 = 100

And PQ2 + RP2 = 72 + 62

= 7 × 7 + 6 × 6

= 49 + 36 = 85

Clearly, QR2 ! PQ2 + RP2

So, the given measures will not forma triplet.

(iii) Let in a triangle XYZ;

XY = 1.5,

YZ = 2 and ZX = 2.5

Here, ZX2 = 2.52 = 2.5 × 2.5

= 6.25


And XY2 + YZ2 = 1.52 + 22

= 1.5 × 1.5 + 2 × 2

= 2.25 + 4 = 6.25

Clearly, ZX2 = XY2 + YZ2

So, the given measures will form atriplet and the right angle is at thevertex Y, i.e., opposite to side oflength 2.5.

7. Let AB and CD be the two poles. DrawAM perpendicular to CD from A tomeet CD at M.

∴ AM = BD, MD = AB = 15 m,

CM = CD – MD = 30 – 15 = 15 m,

And Δ AMC is a right triangle right atM.

∴ AC2 = AM2 + CM2


or 392 = AM2 + 152

or AM2 = 392 – 152 = 39 × 39 – 15 × 15

= 1521 – 225 = 1296 = 36 × 36

∴ AM = 36 m

∴ BD = 36 m (∵ AM = BD)

So, the distance between the feet ofthe poles is 36 m.

8. (i)Given triangle is a right triangle (seefigure).

∴ x2 = 32 + 42


= 9 + 16 = 25 = 5 × 5

∴ x = 5 cm.

(ii) Given triangle is a right triangle

∴ x2 + 92 = 152

or x2 = 152 – 92

= 15 × 15 – 9 × 9

= 225 – 81 = 144

= 12 × 12

∴ x = 12 cm.9. (i) AB + BC = 7 cm + 24 cm = 31 cm

Here, AB + BC > CAi.e., sum of lengths of two sides >length of third side.So, the given measures are the sidesof a triangle ABC.

(ii) PQ + QR = 3 cm + 4 cm = 7 cm

Here, PQ + QR > PR

i.e., sum of lengths of two sides >length of third side.

So, the given measures are the sidesof a triangle PQR.

10. (i) Sum of angles = ∠A + ∠B + ∠C= 40° + 50° + 80°= 170°

Since, the sum of angles is not 180°,therefore, the given angles cannotform a triangle.

(ii) Sum of angles = ∠A + ∠B + ∠C= 75° + 85° + 35°= 195°.

Since, the sum of angles is not 180°,therefore, the given angles cannotform a triangle.

11. ∠A + ∠B + ∠ACB = 180°


or x + 3x + 60° = 180°

or 4x = 180° – 60° = 120°

∴ x = 120

4

ο= 30°

87AIRT GN EL S

∴ 3x = 3 × 30° = 90°

So, all three angles are 30°, 90° and60°.

WORKSHEET –50

1. (i) Angles x and 70° (see figure) arevertically opposite angles.

∴ x = 70°

Now x + y + 30° = 180°


or 70° + y + 30° = 180°

(∵ x = 70°)

∴ y = 180° – 100°

= 80°.

(ii) y + 30° + 100° = 180°


∴ y = 180° – 130° = 50°

Further, x = y + 30°


or x = 50° + 30° = 80°.

2. (i) x is the hypotenuse of the givenright-angled triangle.

∴ x2 = 152 + 82


= 15 × 15 + 8 × 8

= 225 + 64 = 289 = 17 × 17

∴ x = 17.

(ii) In right ΔABC, BC2 + AC2 = AB2


or x2 + 122 = 152

∴ x2 = 152 – 122

= 225 – 144

= 81= 9 × 9

∴ x = 9

Similarly, in right ΔADC,

y2 = 152 – 122 = 9 × 9

∴ y = 9.

3. (i) In ΔXYZ

∠X + ∠Y + ∠Z = 180°


or 60° + 60° + ∠Z = 180°

∴ ∠Z = 180° – 120°

= 60°.

(ii) In ΔLMN,

∠L + ∠M + ∠N = 180°

or 90° + 45° + ∠N = 180°

∴ ∠N = 180° – 135°

= 45°.

4. Let the interior opposite angles are 2xand 3x.

We know that one exterior angle of atriangle is equal to the sum of interioropposite angles.

∴ 125° = 2x + 3x

⇒ x = 125

5°

= 25°

∴ 2x = 2 × 25° = 50°

and 3x = 3 × 25° = 75°

Now, the third angle of the triangle

= 180° – (50° + 75°)


= 180° – 125° = 55°

Thus, all the angles of the triangle areof measure 50°, 75° and 55°.

5. (i) y is an exterior angle and its interioropposite angles are of measures 30°and 60°.

∴ y = 30° + 60°


= 90°.(ii) In ΔABC, AB = AC


∴ ∠ACB = ∠ABC = x (say)Now 40° + x + x = 180° (Angle sum property)

∴ 2x = 180° – 40° = 140°

∴ x = 140

2

ο = 70°

Now, y = 40° + x


= 40 + 70° = 110°.

6. (i) Let the measure of the third anglebe x.

59° + 45° + x = 180° (Angle sum property)or 104° + x = 180°∴ x = 180° – 104° = 76°.

(ii) Let the measure of the third anglebe y.

35° + 116° + y = 180°

or 151° + y = 180°

∴ y = 180° – 151° = 29°.

7. Let the required angle be of measure x.

Since, the triangle is a right triangle,therefore, one of its angles is ofmeasure 90°.

Now, 53° + 90° + x = 180°


or 143° + x = 180°

∴ x = 180° – 143° = 37°.

8. (i) Let ∠ABC = y

Now x = y


Using angle sumproperty of atriangle, we get

x + y + 80° = 180°or x + x + 80° = 180° (∵ x = y)or 2x = 180° – 80° = 100° ∴ x = 50°.

(ii) ∠ABC = ∠ACB = y (say) (∵ AC = AB)

y + 120° = 180° (Linear pair)∴ y = 180° – 120° = 60°.Now 120° = x + y

(Exterior angle

property)

∴ x = 120° – 60° = 60° .

(iii) Let ∠A = y

Here x = y

(∵ AB = BC)

Now, x + y + 90° = 180°


∴ 2x = 180° – 90° = 90° (∵ x = y)

x = 45°.(iv) ∠C = ∠A = 20° (∵ AB = BC)

Now,x + 20° + 20° = 180°


∴ x = 180° – 40° = 140°.

9. Let one of equal angles be x.

Then third angle = x + 30°

So, x + x + x + 30° = 180

or 3x = 150 or x = 50°

Thus required angles are of measures50°, 50° and 80°.

❏❏

89GNOC UR NE EC

WORKSHEET – 51

1. (A) If line segments are congruent, thentheir lengths are equal.

∴ AB = CD .

2. (C) The symbol of congruence ingeometry is { only.

3. (B) Measures of congruent angles areequal.

∴ m ∠A = m ∠B

or m ∠A – m ∠B = 0.

4. (C) If ∆ABC { ∆QRP, then

∠A = ∠Q, ∠B = ∠R and ∠C = ∠P.

Therefore, ∠C corresponds to ∠P.

5. (D) If ∆ PQR { ∆ ABC, then

∠P = ∠A, ∠Q = ∠B and ∠R = ∠C.

6. (B) AAA is not a criterion for congrue-nce of two triangles.

7. (D) Observing the figures, we obtain

AB = PR, BC = RQ and CA = QP.

So, by SSS criterion for congruence,

∆ ABC { ∆ PRQ.

8. (A) ∵ ∠1 = ∠2, ∠3 = ∠4 and includedside AC is common.

∴ ∆ABC { ∆ADC. (ASA criterion).

9. (C) AP = DP, ∠APB = ∠DPC, PB = PC

So, by SAS congruence criterion

∆ APB { ∆ DPC.

10. (A) ∆ LMN { ∆ JKT

⇒ LM = JK, MN = KT, LN = JT.

11. (A) From the given figures, we have

8Chapter

CONGRUENCE

AB = DE, BC = EF, AC = DF

So, we will use SSS criterion forcongruence.

12. (B) Observing the given figures, weget

∠B and ∠P are right angles such that∠B = ∠P

AC and RQ are hypotenuses such thatAC = RQ

AB and RP are sides such thatAB = RP

So, here criterion for congruence isRHS.

13. (D) ∠D = ∠E = 55°

⇒ ∠F = 180° – 55° – 55° = 70°

∠R = ∠F = 70°.

14. (B) ∵ ∆ LMN { ∆ XYZ

∴ LN = XZ15. (C) Since measures of congruent angles

are equal, so, the measure of the otherone will also be 75°.

16. (A) If the radii of two circles are equal,then they are congruent.

17. (D) Observing the figure, we get bySAS criterion of congruence

∆ PQR { ∆ CBA

18. (A) Remaining angle of ∆ ABC

= 180° – 95° = 85°

Now, the measure of required anglewhich corresponds to this angle isequal to 85°.


WORKSHEET –52

1.1.1.1.1. (i)∵ PQ = SR and PS = QR

∴ PQ { SR and PS { QR

(ii) Since AB = BC = CD = DA

AB, BC, CD and DA are congruentto one another.

(iii)∵ CD = DE = EC

So CD , DE and EC are congruentto one another

(iv)∵ NM = NO

∴ NM { NO

2.2.2.2.2.∵ AB { CD

∴ AB = CD

Adding BC to both sides, we get

AB + BC= BC + CD

AC = BD

Since the lengths of line segments ACand BD are equal, so they will becongruent.

Therefore, AC { BD is true.

3.3.3.3.3.∵ AB { CD

∴ AB = CD

∴ CD = 8 cm. (∵AB = 8 cm)

4.4.4.4.4. (i) In ∆ABC and ∆PQR;

∠A = ∠P, AC = PR and ∠C = ∠R

So, by ASA congruence criterion,

∆ ABC { ∆ PQR.

(ii) In ∆ABC and ∆PQR;

∠B = ∠Q = 90°,

hypotenuse CA = hypotenuse PR

and side BC = side QP

So, by RHS congruence criterion,

∆ ABC { ∆ RQP

(iii) In ∆ ABC and ∆ PQR;

∠A = ∠P, AC = PR and ∠C = ∠R

So, by ASA congruence criterion,

∆ ABC { ∆ PQR

Also, ∆ ABC { ∆ RQP

(iv) In ∆ABC and ∆PQR;

AB = QR , BC = RP

and CA = PQ

So, by SSS congruence criterion

∆ABC { ∆QRP.

5.5.5.5.5. (i) In ∆ABC and ∆DEF;

∠B = ∠E = 90°, CA = DF and

BC = ED .

So, ∆ ABC { ∆ FED

(RHS congruence criterion)

(ii) In ∆ LMO and ∆ MNO;

∠MOL = ∠MON = 90°,

ML = MN and OL = ON.

So, ∆ LMO { ∆ NMO. (RHS congruence criterion)6.6.6.6.6. (i) ∆PRM { ∆QRM

(SSS congruence criterion)∴ RP = RQ. (CPCT)

(ii) ∆ MNR { ∆ SQR (SSS congruence criterion)

RM = RS. (CPCT)(iii) ∆ AOB { ∆ DOC

(SAS congruence criterion)∴ OB = OC. (CPCT)

(iv) ∆ ABC { ∆ PQR (RHS congruence criterion)

∴ AC = PR. (CPCT)

91GNOC UR NE EC

7. In ∆PQR and ∆PSR,

∠PRQ = ∠PRS (Each 90°)

PQ = PS

QR = SR

So, ∆ PQR { ∆ PSR (RHS congruence criterion)

8. Join AD.

In ∆ABD and ∆ACD,

AB = AC (Given)

BD = CD

(D is mid-point of BC)

AD = AD (Common)

So, by SSS congruence criterion,

∆ ABD { ∆ ACD.

9. In ∆PQR and ∆PMR,

∠Q = ∠M (Each 90°)

PR = PR (Common)

PQ = PM (Given)

∴ ∆PQR { ∆PMR

(RHS congruence criterion)

So, QR = MR. (CPCT)

WORKSHEET –53

1. In ∆ABC and ∆PQR,

∠B = ∠Q (Each 90°)

Hypotenuse AC = Hypotenuse PR

(Given)

Side BC = Side QR (Given)

So, by RHS congruence criterion, wehave

∆ ABC { ∆ PQR.

2. According to the given conditions, itis clear that ‘the three sides of onetriangle are equal to the three

corresponding sides of anothertriangle”. Then the triangles arecongruent by SSS congruence criterion.

i.e., ∆ ABC { DEF.

(SSS congruence criterion)

3. (i) In ∆ABD and ∆BAC,

∠BAD = ∠ABC (Each 90°)

BD = AC (Given)

AB = BA. (Common side)

(ii) The three pairs of equal partsobtained in part (i) satisfy theconditions of RHS criterion ofcongruence for ∆ BAD and ∆ ABC.

i.e. ∆ BAD { ∆ ABC

∴ ∠D = ∠C

(Corresponding partsof congruent triangles are equal)

(iii) From the part (ii),

∆ ABD { ∆ BAC

∴ AD = BC

(Corresponding partsof congruent triangles are equal)

4. (i) In ∆ABD and ∆CBD,

∠A = ∠C (Each 90°)

Hypotenuse AD = Hypotenuse CD

(Given)

Side AB = Side BC. (Given)

(ii) The three pairs of equal partsobtained in Part (i) satisfy theconditions of RHS criterion ofcongruence for ∆ABD and ∆CBDunder the correspondence

ABD ↔ CBD

∴ ∆ ABD { ∆ CBD.

(iii) ∆ABD { ∆CBD [From part (ii)]

Since corresponding parts of


congruent triangles are equal.

∴ ∠ABD = ∠CBD.

Therefore, BD bisects ∠ABC.

5. In ∆AOC,

∠A + ∠C + ∠AOC = 180°


∠A = 180° – 100° – 20° = 60°

Similarly, in ∆BOD

∠B = 180° – 20° – 60° = 100°

Now, in ∆AOC and ∆BOD,

∠A = ∠D = 60°

AC = BD = 4 cm

∠C = ∠B = 100°

∴ ∆ AOC { ∆ DOB

(ASA congruence criterion)

WORKSHEET –54

1. (i) In ∆LMN, LN = MN

∴ ∠LMN = ∠L

(Angles opposite to equal sides are equal)

Further ∠LMN + ∠L + ∠N = 180°


or 2∠L = 180° – 33° = 147°

(∵∠N = 33° )

∴ ∠L = 147

2

ο = 73.5°

Now x = ∠N + ∠L


= 33° + 73.5° = 106.5°.

(ii) In the given figure, ∠QPR and 70°form a linear pair

∴∠QPR = 180° – 70° = 110°

∵ ∠Q and ∠PRQ are opposite toequal sides

∴ ∠Q = ∠PRQ

Now ∠Q + ∠PRQ + ∠QPR = 180°


or 2∠Q + 110° = 180°

∴ ∠Q = 70

2

ο = 35°

x = ∠QPR + ∠Q = 110° + 35° = 145°.


2. (i) In ∆ABD, AD = BD

∴ ∠1 = 57° (Angles opposite equal sides)

Further AB y DC and BD is transve-rsal.

∴ ∠2 = ∠1 = 57°

(Alternate interior angles)

∠3 = 180° – 57° – 57°

(Angle sum property for ∆ABD)

= 66°

AD y BC and BD is transversal

∴ x = ∠3 = 66°


y = 180° – 57° – 66° = 57°

(Angle sum property for ∆ BCD)

(ii) MN y QR and PQ is transversal

∴ ∠1 = 70° (Corresponding angles)

∠1 + x + 40° = 180°

(Angle sum property for ∆MNP)

or 70° + x + 40° = 180°

93GNOC UR NE EC

∴ x = 180° – 110° = 70°

3. (i) AB = CD, ∠ABE = ∠DCE, BE = CEThus, two sides and the angleincluded between them of ∆ABE areequal to two corresponding sidesand the angle included betweenthem of ∆CDE. So, ∆ABE and ∆CDEare congruent under the correspon-dence. ABE ↔ DCE∴ ∆ ABE { ∆ DCE

(ii) ∠A = ∠E, AC = EF, ∠C = ∠FThus, two angles and the includedside of ∆ABC are equal to twocorresponding angles and theincluded side of ∆EFD.So, ∆ ABC and ∆ EFD are congruentunder the correspondence. ABC ↔ EDF∴ ∆ ABC { ∆ EDF.

4. Since, AP and BQ both are perpendi-culars on AB∴ AP y BQ.Further AP y BQ and PQ is transversal.∴ ∠P = ∠Q (Alternate interior angles) AP= BQ (Given) ∠A = ∠B (Each 90°)∴ ∆ APO { ∆ BQO. (ASA congruence criterion)So, AO = BO (CPCT)⇒ O is the mid-point of ABAnd PO = QO (CPCT)⇒ O is the mid-point of PQ.

5. (i) ∠C = 180° – (30° + 70°) = 80°

(Angle sum property for ∆ ABC )

∠B = ∠E = ∠Q = 30°

BC = EF = QP = 2 cm

∠C = ∠F = ∠P = 80°

So ∆ABC, ∆DEF and ∆PQR arecongruent under the correspondenceABC ↔ DEF ↔ RQP by ASAcongruent criterion.

(ii) In ∆MPQ,

∠P = 180° – (60° + 45°)

= 75°

In ∆ SBD and ∆ QMP,

∠S = ∠P = 75°

SD = MN = 8 cm

∠D = ∠M = 45°

∴ ∆ SDB { ∆ PMQ


In ∆ COG and ∆ KLR,

∠C = ∠L = 60°

CO = LR = 8 cm

∠O = ∠R = 45°

∴ ∆ COG { ∆ LRK.


6. In ∆ABC and ∆ CDE,BC = CD (Given)

∠BCA = ∠DCE (Vertically opposite angles)

AC = EC (Given)So, by SAS congruence criterion,

∆ ABC { ∆ EDC.7. In ∆ ABC and ∆ ADC,

BC = AD (Given)∠BCA = ∠CAD



AC = AC (Common)∴ ∆ABC { ∆CDA

(SAS congruence criterion)So, AB = DC. (CPCT)

WORKSHEET –55

1. If the hypotenuse and one side of aright triangle are respectively equal tothe hypotenuse and one side of anotherright angled triangle, the triangles arecongruent.The given ∆ABC and ∆PQR may becongruence under the followingcorrespondences: A P, B Q, C R A P, B R, C Q

2. (i) ∆ABD { ∆CDB by SAS congruencecriterion.Here corresponding parts are:

AB = CD∠ABD = ∠CDB

BD = DB .(ii) ∆ ABC { ∆ RPQ by ASA congruence

criterion.Here, corresponding parts are:

∠B = ∠PBC = PQ∠C = ∠Q.

(iii) ∆ABC { ∆EDF by SAS congruencecriterionHere, congruence parts are:

AB = ED∠A = ∠ECA = FE .

3. (i) In ∆ ABC and ∆ PQR,

AC = PR = 50 cm

∠C = ∠P = 130°

BC = PQ = 60 cm

∠A = ∠R = 27°

Thus, ∆ABC and ∆PQR are

congruent by SAS as well as ASAcongruence criterion under thecorrespondence ABC ↔ RQP.

(ii) In ∆ABC and ∆PQR,∠A = ∠P = 20°AC = PR = 3.5 cm∠C = ∠R = 70°

Thus ∆ABC and ∆PQR arecongruent by ASA congruencecriterion under the correspondenceABC ↔ PQR.

4. AB y CD and BC is transversal∴ ∠B = ∠C …(i)

(Alternate interior angles)AB y CD and AD is transversal∴ ∠A = ∠D …(ii) (Alternate interior angles)

AB = CD …(iii) (Each 3 cm)From equations (i), (ii) and (iii), weconclude that ∆ CDO { ∆ BAO


5. ∆PRQ and ∆LMN are congruent; andPQ = LM, RQ = NM.

So, ∠Q must be equal to ∠M to becongruent for the triangles. Therefore,∆PRQ and ∆LMN are congruent underthe correspondence

PRQ ↔ LNM

∴ NL = RP = 3 cm and ∠L = ∠P = 30°.

6. In ∆AOC and ∆BOD,

AO = BO (Given)∠AOC = ∠BOD

(Vertically opposite angles)CO = DO (Given)

95GNOC UR NE EC

∴ ∆AOC { ∆BOD

(SAS congruence criterion)

Then AC = BD (CPCT)

∠ACO = ∠BDO (CPCT)

These are alternate interior anglescorresponding to the lines AC and BDwith CD as transversal.

So, AC y BD.

7. In ∆ABC and ∆DCB,

∠BAC = ∠BDC = 90° (Given)

Hypotenuse BC = Hypotenuse BC

(Common)

Side AC = Side DB (Given)

So by RHS congruence criterion, wehave

∆ABC { ∆DCB.

WORKSHEET –56

1. (i) In ∆ABC,

∴ ∠ACB = 60° (∵ AB = AC)

∠ACD = 180° – ∠ACB = 120°


∠D = ∠CAD (∵ AC = CD)

Now 2∠D = 180° – ∠ACD

= 180° – 120° = 60°

∴ ∠D = 60

2

ο = 30°.

x = 180° – (∠B + ∠D)

= 180° – (60° + 30°) = 90°.

(ii) In ∆ABD, AD = BD

∴ ∠2 = 50°

∠1 = 180° – 50° – 50° = 80°

AD y BC and DB is transversal

∴ x = ∠1 = 80°


AB y DC and DB is transversal

∴ ∠3 = ∠2 = 50°


Now, in ∆BCD,

y = 180° – (x + ∠3)

= 180° – (80° + 50°) = 50°.

(iii) In ∆ABC,

∠1 = ∠2 (∵ AC = AB)∠1 + ∠2 + 40° = 180°

or ∠1 + ∠1 + 40° = 180°

∴ ∠1 = 140

2

ο = 70°

Now x = ∠1 + 40° = 70° + 40°


= 110°.

(iv) DE y BC and AB is transversal

∴ ∠1 = 60°

(Corresponding angles)

In ∆ADE,

x + 30° + 60° = 180°

(Angle sum

property)

∴ x = 180° – 90° = 90°.

(v) ∠1 = ∠2 (∵ CA = BA)


∠1 + ∠2 + 30° = 180°


or 2∠1 = 180° – 30° = 150°

∴ ∠1 = 75° = ∠2

Now, x = ∠2 + 30° = 75° + 30°


= 105°

Similary, y = ∠1 + 30° =105°.

2. (i) Observing the figures, we concludethat three sides of one triangle areequal to the three correspondingsides of the other triangle. So, thetriangles are congruent by SSScriterion.

(ii) Observing the figures, we concludethat two sides and the angleincluded between them of onetriangle are equal to correspondingsides and the angle includedbetween them of the other triangle.So, the triangle are congruent bySAS criterion.

(iii) Observing the figures, we concludethat two angles and their includedsides of a triangle will be equal tothe two corresponding angles andthe included side of another triangle.So, the triangles are congruent byASA criterion.

(iv) Observing the figures, we concludethat the hypotenuse and one side ofa right-angled triangle are respecti-vely equal to the hypotenuse andone side of the other right-angledtriangle. So, the triangles arecongruent by RHS criterion.

3. In ∆ ABD and ∆ ACD,BD = CD = 3 cm

∠BDA = ∠CDA = 90°AD = AD (Common)

So, ∆ ABD { ∆ ACD (SAS congruence criterion)

4. (i) ∠1 = x

(Angles opposite

to equal sides)

∠1 + 120° = 180°

(Linear pair)

x = 180° – 120° = 60°

(ii) x = y

(Angles opposite

to equal sides)

105° = x + y


or 105° = x + x

∴ x = 105

2

ο = 52.5°.

(iii) Angles x and y areopposite to equalsides.∴ x = yAngles y and 40°are vertically opposite angles.∴ y = 40°Therefore, x = 40°.

(iv) Angles x and y are opposite to equalsides in the triangle.∴ x = yNow x + y + 90° = 180°(Angle sum property

for the triangle)or x + x + 90° = 180°

∴ x = 902

ο= 45°.

❏❏

97PMOC RA NI QG NAU IT IT E S

WORKSHEET –57

1. (A) 5 km500 m

= 5 1000 m

500 m×

(∵ 1 km = 1000 m)

= 500 10

500×

= 101

.

2. (C)11 m50 cm

= 11 100 cm

50 cm×

(∵ 1 m = 100 cm)

= 11 50 2

50× ×

= 221

.

3. (A) 700 paise

8`=

700 paise8 100 paise×

(∵ ` 1 = 100 paise )

= 78

.

4. (D) 23

= 23

× 33

= 2 33 3××

= 69

.

5. (B) Distance = 16020

× 25 km

= 8 × 25 km = 200 km.

6. (C) Per cent form of 33

= 33

× 100%

= 1 × 100%

= 100%.7. (B) Per cent form of 0.099

= 0.099 × 100%

= 9.9%.

8. (C) Per cent voters who voted ‘yes’

= 80125

× 100%

= 80 × 45

% = 64%.

9. (C) The whole figure is divided into 8equal parts.

Number of shaded parts = 2

∴ Per cent of shaded parts

= 28

× 100% = 25%.

10. (C) 150% = 150100

= 50 350 2

××

= 32

.

11. (D) 11% = 11100

= 0.11.

12. (A) Profit = ` (9200 – 8000) = ` 1200

Profit percentage

= Profit

CP × 100%

= 12008000

× 100% = 120

8%

= 15%.

13. (C)Given ratio is: 1 : 6 : 7.

Their sum = 1 + 6 + 7 = 14

Their percentage are 1

14 × 100%,

614 × 100% and

714 × 100% respectiv-

ely.

i.e., 507 %,

3007 % and

1002 %

i.e., 717 %, 42

67 and 50%.

9Chapter

COMPARING QUANTITIES


14. (C) Interest = Principal × Rate ×Time

100

∴ 560 = 14000 Rate × 2

100×

⇒ Rate = 560 10014000 2

××

% = 5628

%

= 2% per annum.

15. (A) ∵ Out of 125 students, number of absentees = 20

∴ Out of 100 students, number ofabsentees

= 20

125 × 100 = 16

Thus, 16% students are absent.

16. (C) 7% = 7

100 = 0.07.

17. (D) 22% = 22

100 =

1150

.

WORKSHEET –58

1. (i)∵ 189

= 21

∴ 18 : 9 = 2 : 1.

(ii) ∵ 1144

= 14

∴ 11 : 44 = 1 : 4.

(iii) ∵ 101000

= 1

100∴ 10 : 1000 = 1 : 100.

(iv) ∵ 124

= 31

∴ 12 : 4 = 3 : 1.

2. (i) ∵ 10 kg230 g

= 10 1000 g

230 g×

= 1000

23 (∵ 1 kg = 1000 g)

So, ratio of 10 kg to 230 g is 1000 : 23.

(ii) ∵30 days36 hours

= 30 24 hours

36 hours×

= 201

(∵ 1 day = 24 hours)

So, ratio of 30 days to 36 hours is20 : 1.

(iii) ∵ 3 km300 m =

3 1000 m300 m×

= 101

(∵ 1 km = 1000 m)

So, ratio of 3 km to 300 m is 10 : 1.

(iv) ∵ 1 km250 m

= 1000 m250 m

= 41

(∵ 1 km = 1000 m)

So, ratio of 1 km to 250 m is 4 : 1.

3. Meera solves 28 sums in 4 hours

Meera’s speed

= Number of sums

Number of hours taken

= 284

= 7 sums per hour

Shabnam solves 36 sums in 8 hours

Shabnam’s speed

= Number of sums

Number of hours taken

= 368

= 4.5 sum per hours

So, Meera has more speed.

4. Cost of one chair = Cost of 32 chairs

32

= `448032

= ` 140.

(i) Cost of 45 chairs

= 45 × cost of 1chair

= 45 × ` 140

= ` 6300.


(ii) Number of required chairs

=8400

Cost of 1 chair

= 8400140

= 60 chairs.

5. (i)35

= 35

× 100% = 3005

% = 60%.

(ii)58

= 58

× 100% = 5008

% = 62.5%.

(iii) 920

= 9

20 × 100% =

90020

% = 45%.

(iv)54

= 54

× 100% = 500

4% = 125%.

6. First Student:

Fraction = Marks obtained

Maximum marks=

1220

=35

Percentage = 35

× 100% = 3 × 20%

= 60%

Second Student:

Fraction = Marks obtained

Maximum marks=

1620

= 45

Percentage =45

× 100% = 4 × 20%

= 80%.

7. (i) 75% = 75

100 =

34

(ii) 10% = 10

100 =

110

(iii) 1212 % =

252

% = 25

2 100× = 18

(iv) 40% = 40

100 =

410

= 25

.

8. Expenditure on grocery items

= 25% of ` 30000

= ` 30000 × 25

100

= ` 300 × 25

= ` 7500

Expenditure on house rent

= 20% of ` 30000

= ` 30000 × 20

100

= ` 300 × 20 = ` 6000

So Mrs. Shah spends on both

= ` 7500 + ` 6000

= ` 13500.

WORKSHEET –59

1. (i) 30% of 300 kg = 300 × 30

100 kg

= 3 × 30 kg = 90 kg.

(ii) 25% of 25 marks = 25 × 25

100 marks

= 254

marks

= 6.25 marks.

(iii) 50% of 12.30 = 12.30 × 50

100

= 12.30

2 = 6.15.

(iv) 55% of 330 = 330 × 55

100 =

33 5510×

= 1815

10 = 181.5.


2. Number of students who like eatingpizza = 8% of 25

= 25 × 8

100 = 8

4 = 2

So, the number of students who donot like eating pizza = 25 – 2 = 23.

3. Let the total number of students be x.

Number of students passed in Hindi +Number of students passed in English– Number of students passed in both.

= Total number of students.

∴ x × 90

100 + x ×

85100

– 150 = x

or 90x + 85x – 15000 = 100x

or 75x = 15000

∴ x =15000

75or x = 200

Thus, the total number of students is200.

4. Profit = 20% of 1200 = 1200 × 20

100

= 12 × 20 = ` 240

∴ SP = ` 1200 + ` 240 = ` 1440.

5. Principal, P = ` 20000

Rate of interest, R = 5%

Time, T = 2 years

Interest = P × R × T

100 =

20000 5 2100× ×

= 200 × 10 = ` 2000.

6. (i) ∵ 2 km400 m

= 2 1000 m

400 m×

= 204

= 51

∴ 2 km : 400 m = 5 : 1.

(ii) ∵ 1

100 mll

= 1000 ml100 ml

= 101

∴ 1 l : 100 ml = 10 : 1

(iii) ∵ 8

80 paise`

= 8 100 paise

80 paise×

= 101

∴ ` 8 : 80 paise = 10 : 1.

7. 1 : 2 = 12

= 12

× 22

= 24

2 : 3 = 23

∵ 24

! 23∴ 1 : 2 ! 2 : 3

Therefore, 1 : 2 and 2 : 3 are notequivalent.

8. (i) 39% of 800 kg = 800 × 39

100 kg

= 8 × 39 kg = 312 kg.

(ii) 25% of 200 marks = 200 × 25100

marks

= 2 × 25 marks

= 50 marks.

(iii) 53% of 12.30 = 12.30 × 53

100

= 1230100

×53

100

= 65191000

= 6.519.

(iv) 93% of 560 = 560 × 93

100

= 56 93

10×

= 520810

= 520.8.

9. T = 4 years, R = 12%, P = ` 1800

Interest, I = P R T100

= 1800 12 4100× ×

= 18 × 48 = ` 864Amount = P + I = ` 1800 + ` 864

= ` 2664.


10. ∵ Out of 500 fruits, number of rottenfruits = 5

∴ Out of 1 fruit, number of rotten fruits

= 5500

= 1

100

∴ Out of 100 fruits, number of rottenfruits

= 1

100 × 100 = 1

So, 1% of fruits is rotten.

WORKSHEET –60

1. (i) 25% of 64 = 64 × 25

100

= 64 × 14

= 16.

(ii) 1212

= 12 2 1

2× +

= 252

1212

% of 900 = 900 × 25200

= 9 25

2×

= 2252

= 112.5.

(iii) 2% of 2 hours

= 2 × 2

100 hours

= 4

100 hours = 0.04 hours.

= 0.04 × 60 minutes = 2.4 minutes.

2. Let the whole quantity be x km.

50% of x = 1000

or x × 50

100= 1000

or x = 2000 km.

3. The whole figure is divided into 8equal parts.

The number of shaded parts = 4.

∴ Percentage of the shaded portion

= 48 × 100% =

1002 % = 50%.

4. Number of all balls = 800

Number of blue balls = 480

∴ Percentage of blue balls

= Number of blue ballsNumber of all balls

× 100%

= 480800

× 100% = 4808

% = 60%

5. (i) ∵ 20

40 paise`

= 20 100 paise

480 paise×

= 20048

= 256

∴ ` 20 : 480 paise = 25 : 6.

(ii) ∵ 5 km600 m

= 5 1000 m

600 m×

= 506

= 253

∴ 5 km : 600 m = 25 : 3.

(iii) ∵ 3

1500 mll

= 3000 ml1500 ml

= 3015

= 21

∴ 3l : 1500 ml = 2 : 1.

6. (i) The selling price (SP) of the book ismore than the buying price (CP).So, the book provides profit.

Profit = SP – CP

= ` 250 – ` 200 = ` 50

Profit % = Profit

CP × 100%

= 50200

× 100% = 25%.


(ii) The SP of the chair is more than theCP. So, the chair provides profit.

Profit = SP – CP

= ` 18500 – ` 15000

= ` 3500

Profit % = Profit

CP × 100%

= 3500

15000 × 100%

= 35015

% = 703

% = 2313 %.

7. CP = ` 20000, SP = ` 24000

Profit = SP – CP

= ` 24000 – ` 20000

= ` 4000

Profit % = Profit

CP × 100%

= 4000

20000 × 100%

= 402

% = 20%.

8. Number of children who like playingcricket = 16% of 200

= 200 × 16

100 = 32.

Number of children who do not likeplaying cricket = 200 – 32 = 168.

9. (i) 10% of 18 litres

= 18 × 10100

litres = 1810

litres

= 1.8 litres.

(ii) 50% of 18.40

= 18.40 × 50

100 =

18.402

= 9.20.

10. Let the total number of students be x

∴ Number of students who like to take

coffee = x × 72

100 =

72100

x

And number of students who like to

take tea = x × 52

100 =

52100

x

Now,72100

x +

52100

x – 144 = x

Multiplying throughout by 100, we get72x + 52x – 14400 = 100x

or 124x – 100x = 14400

or 24x = 14400

or x = 14400

24= 600

Thus, the total number of students inthe group is 600.

WORKSHEET – 61

1. Profit = 12% of CP = ` 50 × 12

100 = ` 6

SP = CP + Profit = ` 50 + ` 6 = ` 56.

2. Loss is 5% of CP.

∴ CP = SP × 100(100 – 5)

= ` 570 × 10095

= ` 6 × 100 = ` 600.

3. T = 5 years, P = ` 2500, R = 3%

I = PRT100

= 2500 3 5

100× ×

= 25 × 15

= ` 375

Amount = P + I = ` 2500 + ` 375

= ` 2875.

4. I = ` 8100, P = ` 72000, T = 3 years,

R = ?


I = PRT100

or R = I × 100

P T =

8100 10072000 3

××

= 81 1072 3×× =

9 108 3×× =

308

= 3.75

So, the rate of interest is 3.75% perannum.

5. (i) 0.75 = 0.75 × 100%

= 75

100 × 100% = 75%.

(ii) ∵ 214

= 2 4 1

4× +

= 8 1

4+

= 94

∴ 214

= 214

× 100% = 94

× 100%

= 9 × 25% = 225%.

(iii)4750 =

4750 × 100% = 47 × 2% = 94%.

6. Speed =Distance

Time =

1203

= 40 km/h

(i) Distance covered = Speed × Time

= 40 × 8 = 320 km

(ii) Time =Distance

Speed=

84040 = 21 hours.

7. Speed = Distance

Time=

18006

= 300 km/hour.

8. ` 30 = 30 × 100 paise

∴ 30900 paise

`=

30 100 paise900 paise×

= 309

= 103

or ` 30 : 900 paise = 10 : 3.

9. 81

50=

8 50 150

× + =

400 150+

= 40150

= 40150

× 100% = 401 × 2%

= 802%.

10. 16% of 7700 = 7700 × 16

100

= 77 × 16 = 1232.

11. Sum of ratios = 4 + 2 = 6

First part = 46

× 18000 = 4 × 3000

= ` 12000

Second part = 26

× 18000 = 2 × 3000

= ` 6000.

WORKSHEET –62

1. 4 km = 4 × 1000 m = 4000 m

∴ 4 km : 400 m = 4000 m : 400 m

= 10 : 1.

2. ∵ 4 cm = 1000 km

∴ 1 cm = 1000

4 km

∴ 3.5 cm = 1000

4 × 3.5 km

= 3500

4 km = 875 km

So, the actual distance is 875 km.

3. ∵ Cost of 7 chairs = ` 714

∴ Cost of 1 chair = ` 7147 = ` 102

∴ Cost of 83 chairs = ` 102 × 83

= ` 8466.


4.4.4.4.4. Percentage of absent students

= Number of absentees

Total number of students × 100%

= 1045

× 100% = 200

9% = 22

29

%.

5.5.5.5.5. Number of students owning a bicycle

= 55% of 1200.

= 1200 × 55

100 = 660

∴ Number of students not owning abicycle

= 1200 – 660 = 540.

6.6.6.6.6. 150% = 150100

= 1510

= 1.5.

7.7.7.7.7. Whole quantity = 20 × 10080

minutes

= 100

4 minutes

= 25 minutes.

8.8.8.8.8. Number of won matches

= 25% of 20

= 20 × 25100

= 520100

= 5

∴ Number of lost matches

= 20 – No. of won matches

= 20 – 5 = 15.

9.9.9.9.9. 20% of 25 sweets = 20 × 25

100 =

500100

= 5 sweets

80% of 25 sweets = 80 × 25

100 =

2000100

= 20 sweets

Hence, Manu gets 5 sweets and Tanugets 20 sweets.

10.10.10.10.10. Let the angles be 2A, 3A and 4A.

∴ 2A + 3A + 4A = 180° or 9A = 180°

or A = 180

9

ο or A = 20°

∴ 2A = 2 × 20° = 40°, 3A = 3 × 20°= 60° and 4A = 4 × 20° = 80°.

So, the angles are of measures 40°, 60°and 80°.

11.11.11.11.11. Profit = SP – CP

= ` 2300 – ` 2100 = ` 200

Profit % = ProfitCP

× 100%

= 200

2100× 100% =

200%

21

= 91121

%.

12.12.12.12.12.18

= 18

× 100% = 252

% = 1

122 %.

WORKSHEET –63

1.1.1.1.1. Given ratio is 2 : 3 : 5

Sum of those = 2 + 3 + 5 = 10

Percentage of the first part

= 2

10 × 100 = 20%

Percentage of the second part

= 3

10 × 100 = 30%

Percentage of the third part

= 5

10 × 100 = 50%.

2.2.2.2.2. P = ` 8000, R = 3%,

T = 4 months = 4

12 year =

13

year


Interest, I = PRT100

=

18000 3

3100

× ×

= 8000 3 1

3 100× ×× = 80

Thus, ` 80 are to be paid as interest.

3. Percentage of marks secured byNandini

= Marks secured

Maximum marks × 100%

= 2225

× 100% = 88%.

Percentage of marks obtained byBhawna

= Marks secured


= 4350

× 100% = 86%

Nandini secured more percentage ofmarks, and so her performance isbetter.

4. P = ` 2500, R = 4%,

T = 9 months = 9

12 year =

34

year

I = PRT100

= 2500 4 3

4 100× ×× = ` 75

Amount = I + P = ` 75 + ` 2500

= ` 2575

Thus, the interest is ` 75 and amountis ` 2575.

5. Profit = 18% of CP = CP × 18

100

SP = CP + Profit

So, 150 = CP + CP × 18

100

= CP (1 + 18

100 )

= 118100 CP

Therefore, CP = 150 100

118×

= 75 100

59×

= 7500

59

= 127.12

Thus, the cost price is ` 127.12.

6. Decrease = ` 90 – ` 50 = ` 40.

Decrease percentage

= Decrease

Original price × 100%

= 4090

× 100% = 4009

%

= 4449 %.

7. Loss = 10% of CP = CP × 10

100

= 1

10CP

∵ SP = CP – Loss

∴ 270 = CP – 1

10 CP = (1 – 1

10 ) CP

= 9

10CP

∴ CP = 270 10

9×

= 30 × 10 = 300

Therefore, the cost price is ` 300.

8. 33% of ` 13500 = ` 13500 × 33

100

= ` 135 × 33

= ` 4455.


9.20 m80 cm

= 20 100 cm

80 cm×

= 100

4 = 251

∴ 20 m : 80 cm = 25 : 1.

10. We know that 1 dozen = 12

∵ Cost of 1 mango = ` 2.25 = ` 225100

∴ Cost of 1 dozen mangoes

= ` 225100

× 12 = ` 2700100

= ` 27.

11. 1 : 2 or 12

= 12

× 33

= 36

and 3 : 8 or 38

Since 36 !

38

. Therefore, 1 : 2 and 3 : 8

are not equivalent.

12. SP = CP + Profit

= CP + 15% of CP = CP + 15

100CP

= CP ( )151

100+ = CP ×

115100

= 600 × 115100

= 6 × 115 = 690

Therefore, selling price of the book is` 690.

WORKSHEET –64

1. (i) ∵3072 =

6 56 12××

= 5

12

∴ 30 : 72 = 5 : 12.

(ii) ∵ 7.25

10.25 =

7.2510.25

× 100100

= 725

1025

= 25 2925 41

××

= 2941

∴ 7.25 : 10.25 = 29 : 41.

2. (i) 6

70 paise`

= 6 100 paise

70 paise×

= 607

i.e., ` 6 : 72 paise = 60 : 7.

(ii)3 km800 m

= 3 1000 m

800 m×

= 308

= 154

i.e., 3 km : 800 m = 15 : 4.

3. ∵ Distance covered in 3 hours= 120 km

∴ Distance covered in 1 hour

= 120

3 = 40 km

∴ Distance covered in 18 hours= 40 × 18 km

= 720 km.

4. Total number of rings = 20 + 10 = 30

Percentage of gold rings

= Number of gold ringsTotal number of rings

× 100%

= 2030

× 100% = 2003

% = 6623 %

Percentage of silver rings

= Number of silver ringsTotal number of rings

× 100%

= 1030

× 100% = 100

3% = 33

13 %.

5. (i)1725

= 1725

× 100% = 17 × 4% = 68%.

(ii) 114

= 114 × 100% =

54 × 100%

= 5 × 25% = 125%.

6. Percentage of broken eggs

= Number of broken eggs

Total number of eggs × 100%


= 50400

× 100% = 504

% = 12.5%

7. I = ` 2025, T = 3 years, R = 2%.

I = PRT100

or P = I × 100

RT

∴ P = 2025 100

2 3××

= 675 × 50

= 33750.

Therefore, the loan taken was ` 33750.

8.14

= 14

× 100% = 25%.

9. Sum of ratios = 4 + 2 = 6

First part = 46

× 48000 = 4 × 8000

= ` 32000

Second part = 26

× 48000 = 2 × 8000

= ` 16000.

10. Percentage of marks secured byNeelam

= Marks secured


= 2125

× 100% = 84%

Percentage of marks secured by Bina

= Marks securedMaximum marks

× 100%

= 4350 × 100% = 86%.

Since Bina secured more percentage ofmarks, therefore, her performance isbetter.

11. Number of won matches = 50% of 30

= 30 × 50

100 = 15

No. of lost matches = Total no. ofmatches – No. of won matches

= 30 – 15 = 15.

12. Let the angles be 8A, 3A and 7Arespectively.

∴8A + 3A + 7A = 180°

or 18A = 180° or A = 10°

Therefore, 8A = 8 × 10° = 80°,

3A = 3 × 10° = 30°

and 7A = 7 × 10° = 70°

Thus, the values of the angles are 80°,30° and 70°.

WORKSHEET –65

1. Let the parts be 4x, 3x and 5xrespectively.

Sum of those = 4x + 3x + 5x = 12x

Percentage of the first part

= 4

12xx

× 100% = 100

3%

= 3313 %

Percentage of the second part

= 3

12xx

× 100% = 100

4%

= 25%

Percentage of the third part

= 5

12xx

× 100% = 50012

%

= 4123 %.

2. P = ` 5000, R = 15%, T = 2 years

I = PRT100

= 5000 15 2100× × = 50 × 30

= 1500

Amount = I + P = 1500 + 5000 = 6500


Thus, the interest is ` 1500 and theamount is ` 6500.

3. 75% = 75

100 =

34

or 3 : 4.

4. ∵ 70% of a quantity = 35

∴ 1% of it = 3570

∴ 100% of it = 3570

× 100 = 50

Therefore, the whole quantity is ` 50.5. Total CP = ` 20000 + ` 500 = ` 20500

SP = ` 30000∴ Profit = SP – CP

= ` 30000 – ` 20500= ` 9500.

Profit per cent = Profit

CP × 100%

= 9500

20500 × 100%

= 9500205

% = 46.34%.

6. T = 3 years, I = ` 450, R = 5%

I = PRT100

or P = I × 100

RT =

450 1005 3××

= 30 × 100 = 3000

So, the sum is ` 3000.

7. Number of students who got firstdivision

= 75% of 1500

= 75

100 × 1500 = 75 × 15

= 1125

∴ Number of students who did notget first division

= 1500 – 1125 = 375.

8. Let the principal be P,

then amount = 2P

R = 10%

∴ I = 2P – P = P

Now I = PRT100 or T =

I × 100P × R

or T = P × 100P × 10

= 10.

Thus, the required number of years is10.

9. I = ` 4500, P = ` 72000, T = 3 years,

R = ?

I = PRT100

or R = I × 100

PT

or R = 4500 10072000 3

××

= 450

72 3× =

15072

= 2512

or R = 21

12 % per annum.

10. P = ` 18000, R = 18%,

T = 6 months = 6

12=

12

year

I = PRT100

=

118000 18

2100

× ×

= 180 18

2×

= 1620

Amount = I + P = 1620 + 18000

= 19620

Thus, interest = ` 1620,

amount = ` 19620.

11. CP of a table providing proift

= ` 990 × 100110

= ` 900

CP of other table providing loss


= ` 990 × 10090

= ` 1100.

So, the cost prices of the tables arerespectively ` 900 and ` 1100.

12. Let the CP of the table be ` P.

Then P – P × 5100

= 540 or P – P20

= 540

or 19P20

= 540 or P = 540 20

19×

= 10800

19or P = ` 568.42 (approx.)

13. Profit = 20% of CP = 20

100 × CP =

CP5

Since SP = CP + Profit

∴ 720 = CP + CP5

or 720 = 65

CP

or CP = 720 56× = 600.

Hence, the cost price is ` 600.

WORKSHEET –66

1. (i) ∵ 25% of a number = 18

∴ 1% of it = 1825

∴ 100% of it = 1825 × 100

= 72

So, the required number is 72.

(ii) ∵ 75% of a number = 15

∴ 1% of it = 1575

∴ 100% of it = 1575

× 100

= 20

So, the required number is 20.

2. (i) 40% of 24.36

= 40

100 × 24.36 =

40 2436100 100

××

= 9744010000

= 9.744.

(ii) 10% of 69 litres

= 10

100 × 69 litres = 6.9 litres.

(iii) 35% of 980

= 35

100 × 980 =

343010

= 343.

3. (i) 3 parts are shaded out of 8 parts

So, fraction of the shaded part = 38

and percentage of the shaded part

= 38

× 100% = 3712 %.

(ii) 4 parts are shaded out of 8 parts

So, fraction of the shaded part

= 48

= 12

and percentage of the shaded part

= 12

× 100% = 50%.

4. Number of people having cars = 75%Number of people not having cars

= (100 – 75)% = 25%.5. Number of children who like watching

movies = 25% of 80

= 25

100 × 80 =

25 810×

= 20010

= 20.

6. (i) 100% of Raju’s weight

= 7.210 × 100 kg = 72 kg.

So, Raju’s weight is 72 kg.


(ii) 100% of the journey

= 6250 × 100 km = 124 km

So, the whole journey is 124 kmlong.

(iii) ∵ 5% of the enrolment = 75

∴ 100% of the enrolment

= 755 × 100 = 75 × 20 = 1500

So, the strength of the school is 1500children.

(iv) ∵ 30% of the total marks = 60

∴ 100% of the total marks

= 6030 × 100 = 200

So, the total marks of the paper is200.

7. Reeta’s marks in Hindi

= 40% of maximum marks in Hindi

= 40

100 × 150 = 4 × 15 = 60

Reeta’s marks in Maths

= 55% of maximum marks in Maths

= 55

100 × 180 =

99010

= 99.

8. Percentage of students who did notlike playing cricket = (100 – 60)%

= 40 %

And their number = 40 % of 1500

= 40

100 × 1500

= 40 × 15 = 600.

9. (i) Loss = CP – SP

= ` 50 – ` 30 = ` 20

Loss % = LossCP × 100%

= 2050 × 100% = 40%.

(ii) Profit = SP – CP

= ` 2500000 – ` 2225000

= ` 275000

Profit %= Pfofit

CP × 100%

= 275000

2225000 × 100%

= 275002225 % =

110089 %

= 123289 %.

10. Loss% = LossCP × 100%

∴ 10% = LossCP × 100%

or Loss = CP10

Now, Loss = CP – SP

orCP10 = CP – 819 or 819 =

910

CP

or CP = 819 10

9×

= 91 × 10 = 910

So, the cost price was ` 910.

11. Let CP = ` R.

Then, loss = R × 20

100 =

R5

Now SP = CP – Loss

or CP = SP + Loss


∴ R = 13500 + R5

or R – R5

= 13500

or4R5

= 13500 or R = 5 × 3375

or R = 16875

So, the cost price of the article was` 16875.

WORKSHEET –67

1. Let the CP of the cow providing profitbe x1 and the CP of the cow providingloss be x2.

So x1 × 110100

= 1980

and x2 × 90

100 = 1980

or x1 = 1980 10

11×

and x2 = 1980 10

9×

or x1 = 1800 and x2 = 2200

Hence, the cost prices of the cows were` 1800 and ` 2200 respectively.

2. Cost price for me = ` 1275

Profit = 10% of ` 1275 = ` 10

100 × 1275

= ` 127.50

Selling price for me

= ` 1275 + ` 127.50

= ` 1402.50.

3. I = ` 1080, T = 2 years, P = ` 9000,

R = ?

We have I = PRT100

∴ R = I × 100P × T

∴ R = 1080 100

9000 2×× =

10818 = 6

So, the rate of interest is 6% per annum.

4. CP of the machine providing profit

= ` 120000 × 100125

= ` 960 × 100

= ` 96000

CP of the machine providing loss

= ` 120000 × 10075

= ` 1600 × 100

= ` 160000

Total CP = ` 96000 + ` 160000

= ` 256000

Total SP = 2 × ` 120000 = ` 240000

Since CP > SP. So, there is a loss.

Loss = ` 256000 – ` 240000 = ` 16000

Loss per cent = Loss

Total CP × 100%

= 16000256000 × 100%

= 10016 % = 6

14 %.

5. Total CP for shyam

= 45000 + 200 = ` 45200

Profit = Profit% × CP

100

= 10 45200

100×

= ` 4520 .

6. CP of fan = 1200 + 200 = ` 1400

SP of fan to gain 10%

= 1400 + 1400 × 10

100= 1400 + 140 = ` 1540.


7. P = ` 500, I = ` 105, T = 6 years, R = ?

I = PRT100

or R = I × 100

PT

∴ R = 105 100

500 6××

= 216

= 72 = 3.5%.

8. P = ` 300, I = ` 60, R = 5%, T = ?

I = PRT100 or T =

I × 100PR

∴ T = 60 100300 5×× = 4 years.

9. ∵ In 2100 km petrol consumed

= 28 litres

∴ In 1 km petrol consumed

= 28

2100 litre

∴ In 3600 km petrol consumed

= 28

2100 × 3600 litres

= 4 36

3×

litres = 48 litres.

10. (i) ∵ 150 steps = 125 m

∴ 1 step = 125150

m = 56

m

∴ 360 steps = 56 × 360 m

= 300 m

Harsh covers a distance of 300 m in360 steps.

(ii) Required number of steps

= 172.8 m

Distance in 1 step

= 172.8 m

5 m

6

= 1728

10 ×

65 = 207.36

Clearly Harsh takes 207 steps and afraction of 1 step, but the step cannotbe in decimal.

So, he will take 208 steps.

11. Cost of 1 chair = 3532.50

15 = ` 235.50

Number of chairs

= 5416 50

Cost of 1 chair` .

= 5416.50 235.50

``

= 54165023550

= 23.

12. (i) x : 5 : : 28 : 35

or5x

= 2835

∴ x = 5 28

35×

= 287 = 4.

(ii) 16 : x : : x : 25

or16x

= 25x

or x2 = 16 × 25 = 4 × 4 × 5 × 5

∴ x = 4 × 5 = 20.

WORKSHEET –68

1. (i) Ratio of 12 and

14

= 12 :

14

= 2 : 1

Ratio of 17 and

114 =

17

: 114

= 2 : 1

Since 12 :

14 =

17 :

114 .

Therefore, 12 ,

14 ,

17 ,

114 are in

proportion.

(ii) Ratio of 2 and 312 = 2 :

72 = 4 : 7


Ratio of 3 and 412

= 3 : 92 = 6 :9

= 4 : 6

Since 4 : 7 ! 4 : 6

Therefore, 2, 312 , 3, 4

12 are not in

proportion.

2. (i) Let the fourth proportion be x.

Then, 8 : 32 = 217 : x or 832 =

217x .

or14

= 217

x∴ x = 4 × 217 = 868

Thus, the fourth proportion is 868.

(ii) Let the fourth proportion be y.

Then 3 kg : 7 kg = 15 kg : y

or37

= 15y

∴ y = 7 15

3×

= 7 × 5 = 35

Thus, the fourth proportion is35 kg.

3. ∵ Weight of wheat in 12 bags = 90 kg

∴ Weight of wheat in 1 bag

= 9012

kg = 152

kg

∴ Weight of wheat in 20 bags

= 152 × 20 kg = 150 kg.

4. ∵ Cost of 16 books = ` 72

∴ Cost of 1 book = ` 7216 = `

92

∴Cost of 30 books = ` 92

× 30

= ` 135

Required number of books

= 207

Cost of 1 book`

= 207

9 2

`

`

= 207 2

9×

= 23 × 2 = 46.

5. ∵ Distance covered in 450 steps

= 225 m

∴ Distance covered in 1 step

= 225450 m =

12 m

∴ Distance covered in 900 steps

= 12 × 900 m = 450 m.

6. (i) Required percentage

= 25 paise

2` × 100%

= 25 paise

200 paise × 100%

= 252

% = 1

122

%.

(ii) Required percentage

= 75 m1 km

× 100%

= 75 m

1000 m × 100% =

7510 %

= 7.5%.

7. 4.8% = 4.8100

= 481000

= 6

125.

8. (i) Required percentage

= 400 m4 km

× 100%

= 400 m

4000 m × 100% = 10%.


(ii) Required percentage

= 40 kg

1 quintal × 100%

= 40 kg

100 kg × 100% = 40%.

9. (i) 75% of 400

= 75

100 × 400 = 75 × 4 = 300

Decreasing 300 from 400 , we get400 – 300 = 100.

(ii) 5% of 120

= 5

100 × 120 =

600100

= 6

Increasing ` 6 in, we get

` 120 + ` 6 = ` 126.

10. CP = SP + Loss = ` 20 + ` 5 = ` 25

Loss per cent = LossCP

× 100%

= 5

25 × 100% = 100

5%

= 20%.

11. Gain per cent = GainCP

× 100

or Gain = CP × Gain per cent

100

= 20.25 10

100×

= ` 2.025

SP = Gain + CP

= ` 2.025 + ` 20.25 = ` 22.275

= ` 22.28.

12. Number of papaya trees = 40% of 480

= 40100

× 480 = 4 × 48 = 192

Number of other trees = 480 – 192= 288.

WORKSHEET –69

1. CP = Buying price + Transportation charges

= ` 80000 + ` 1000 = ` 81000

SP = ` 50000

Loss = CP – SP

= ` 81000 – ` 50000 = ` 31000

Loss% = LossCP

× 100%

= 31000 81000

``

× 100% = 3100

81%

= 382281 %.

2. SP = CP + CP × Gain%

100

= 200 + 200 × 10100

= 200 ( )101

100+ = 200 × 11

10 = 220

So, the selling price is ` 220.

3. Let CP = x

SP = CP – CP × Loss%

100

∴ 8 = x – x × 20

100 or 8 = x ( )201

100−

or 8 = x × 45 or x =

8 54×

or x = 10

Hence, the cost price is ` 10.

4. Let CP of 1 chair be x

Then CP of 10 chairs = 10x

∴ SP of 16 chairs = 10x

∴ SP of 1 chair = 1016

x =

58x


Since the CP of 1 chair is greater thanthe SP of 1 chair. Therefore, there isloss.

Loss on 1 chair = x – 58x

= 38x

Loss % = Loss

CP of 1 chair × 100%

=

38x

x × 100% =

3 1008×

%

= 3 25

2×

% = 1

372 %.

5. Let CP of 1 article = x

∴ CP of 6 articles = 6x

∴ SP of 4 articles = 6x

∴ SP of 1 article = 64x

= 32x

Gain = SP of 1 article – CP of 1 article

= 32x

– x = 2x

Gain% = Gain

CP of 1 article × 100%

= 2x

x × 100% = 50%.

6. P = ` 8500, R = 18%, T = 3 years

I = PRT100

= 8500 18 3

100× ×

= 85 × 54

= ` 4590

Amount = Principal + Interest

= ` 8500 + ` 4590 = ` 13090.

7. Let principal = x

Then amount = 2x

∴ Interest, I = 2x – x = x

T = 20 years

Now I = PRT100

or R = I × 100

PT

∴ R = × 100 × 20

xx

= 10020

= 5

Thus, the rate of interest is 5% perannum.

8. P = ` 300, Amount = ` 300 × 2 = ` 600,I = Amount – P = ` 600 – ` 300 = ` 300,R = 4%

We have I = PRT100

or T = I × 100P × R

∴ T = 300 100300 4

×× = 25 years.

9. ∵ Cost of 280 match boxes = ` 36

∴ Cost of 1 match box = ` 36280

∴ Cost of 650 match boxes

= ` 36280 × 650 = `

97 × 65

= ` 83.57.

10. Let the third proportion be x.

Then 251

46

= 25x

or 25 6

25×

= 25x

∴ x = 25 25 6

25× ×

= 150.

11. Let the fourth proportion be x.

Then4.81.6 =

5.4x

∴ x = 1.6 5.4

4.8×

= 16 5448 10

××

= 54

3 10× = 1.8.

12.Original price

New price=

35

or New price = 53 × Original price

= 53 × 7500 = 5 × 2500

= 12,5000So, the new price is ` 12,500.

❏❏


WORKSHEET –70

1. (D) 30 45−

= 3045

− =

23

−.

2. (A) 45

= 0.8. It is less than 1.

So, 45

lies to the left of 1.

3. (B) 47 is a positive quantity and – 1 is

a negative quantity

So, 47

> – 1.

4. (B) A number which is greater than – 3and less than – 2 is

– 2.5 or – 2510

or – 52 .

5. (A) LCM of 5, 7 and 9 = 5 × 7 × 9 = 315

45

− =

45

− ×

6363

= 252315

−

67

− =

67

− ×

4545

= 270315

−

– 89

= – 89

× 3535

= – 280315

∵ – 280315

< – 270315

< – 252315

∴ – 89

< – 67

< – 45

So, – 45

is the greatest.

10Chapter

RATIONAL NUMBERS

6. (D) 35

= – 3– 5

! – 5– 3

.

7. (B) Next two numbers in the pattern– 1, – 2, – 3 are – 4 and – 5 respectively.

Next two numbers in the pattern 3, 6,9 are 12 and 15 respectively.

Therefore, the required numbers are

– 412

and – 515 .

8. (C) – 4672

= – 23 236 2

××

= – 2336

.

9. (D) P = 113

= 3.666...

Therefore, P lies between 3 and 4.

10. (A) Since 3, 7, 9 are in the ascendingorder.

Therefore, 35 ,

75 ,

95 are in the

ascending order.

11. (B) LCM of 4, 7, and 8 is 56

– 57

= – 4056

; – 54

= – 7056

; – 58

= – 3556

∵ – 35, – 40, – 70 are in thedescending order.

∴ – 3556 ,

– 4056 ,

– 7056 are in the

descending order.

∴ 58

−,

57

−,

54

− are in the

descending order.

117ITAR NO LA N BMU RE S

12. (A) LCM of 10 and 15 is 30.

910−

– 1115

= 9 3 – 11 2

30− × ×

= 27 22

30− −

= 4930

−.

13. (C) 12

– 13

= 3 2

6−

= 16

.

14. (B) –72

× ( )43

− = 72

× 43

[∵ – a × (– b) = a × b]

= 143

.

15. (C) Reciprocal of 72

− is ( )

1 72

−

= 2 7−

= 27

−.

16. (D)Reciprocal of 1= 11

= 1

The product of 1 and its reciprocal

= 1 × 1 = 1.

WORKSHEET – 71

1. (i) 23

− = – ( )2

3

It is a negative rational number.

(ii)4 5−

= 45

− = – ( )4

5

It is a negative rational number.

2. 27

− =

27

− ×

1111

= 2277

− =

2277

− ×

44

= 88

308−

311−

= 3

11−

× 77 =

2177

−=

2177

− ×

44

= 84

308−

∵ 88

308−

< 87

308−

< 86

308−

< 85

308−

< 84

308−

Therefore, the required three rationalnumbers are:

87308−

, 86

308−

and 84

308−

.

3. First term = 1 8−

Second term = 2 16−

= 2

8 2− ×

Third term = 3 24−

= 3

8 3− ×

Fourth term = 4 32−

= 4 8 4− ×

Similarly, fifth term = 5

8 5− × =

5 40−

sixth term = 6

8 6− × =

6 48−

and seventh term = 7

8 7− × =

7 56−

.

4.

The point P represents 35

− on the

number line.

5. – 312

= – ( )13

2 = 72

−

= 7 52 5

− ×× =

3510

−

– 425

= – ( )24

5 = 225

−


= 22 25 2

− ××

= 4410

−

∵ 3510

− >

4410

−

Therefore, – 312 is greater integer.

6.23

= 2 23 2

××

= 46

∵ – 1 < 1 < 4

∴ 16

− <

16 <

46 or

16

− <

16 <

23

i.e., 16

−,

16 ,

23 are in ascending order.

7. (i)34

= 3 ( 6)4 ( 6)

× −× −

= 18 24

−−

(∵ 244

− = – 6)

So, the required rational number is– 18.

(ii)842 =

421

= 4 ( 1)21 (– 1)

× −×

= – 4– 21

(∵ 2121

− = – 1)

So, the required rational number is– 4.

8. (i) 5

13−

+ ( 2)

13−

= 5 ( 2)

13− + −

(Denominators are same)

= 5 213

− − =

713−

.

(ii) 38

− +

1820

= – 38

+ 9

10

= – 3 × 5 + 9 × 4

40

= – 15 + 36

40 =

2140

.

9. (i) Let us find LCM of 12, 4 and 8.

∴ LCM = 2 × 2 × 2 × 3 = 24

Now, 1

12 + ( ) 3

4−

+ 78

= 2 1 6 ( 3) 3 7

24× + × − + ×

= 2 18 21

24− +

= 5

24 .

(ii) 325

– 710

+ ( ) 215−

– 101

30

= 175

– 710

– 2

15 –

30130

= 6 17 3 (– 7) 2 (– 2) 1 (– 301)

30× + × + × + ×

= 102 – 21 4 301

30− −

= 22430

−

[∵ LCM (5, 10, 15, 30) = 30]

= 11215

− = – 7

715

.

10. (i)6

14 – ( ) 5

7−

= 6

14 +

57 =

6 1014+

= 1614

= 87

= 117

.

(ii)5

16 – ( ) 2

8−

= 5

16 +

28

= 5 416+

= 9

16.

11. (i) 7

12−

× 8 = (7 8)

12− ×

= 5612

− =

143

−

= – 423 .

2 12, 4, 82 6, 2, 42 3, 1, 23 3, 1, 1

1, 1, 1


(ii)4

10 ×

512−

× 25

= 4

12 ×

55−

× 2

10

= 13

× (– 1) × 15

= –1

15.

12.12.12.12.12. 7

12−

÷ 14 = 7

12−

÷ 141

= 7

12−

× 1

14

= 1

12 2−×

= –1

24.

13.13.13.13.13. (i) Reciprocal of 17−

= ( )1 17−

= 7 1−

= –7.

(ii) Reciprocal of 5 11−−

= ( )1 5 11−−

= 11 5

−−

= 115 .

WORKSHEET –72

1.1.1.1.1. (i) In 78−

;

8 is a positive and – 7 is a negative.So, this is a negative rationalnumber.

(ii) In 2 3

−−

; – 2 and – 3 both are negative.

So, this is a positive rational number.

(iii) In 011 ; 0 is neither positive nor

negative and 11 is a positive.So, this rational number is neitherpositive nor negative.

(iv) In 1215 ; 12 and 15 both are positive.

So, the rational number is positive.2.2.2.2.2. Three rational numbers between 3 and

– 3 are 2, 1 and 0.

3.3.3.3.3. (i) 13−

= 1 23 2− ×

× = 26−

13−

= 1 33 3− ×

× =

39−

13−

= 1 43 4− ×

× =

412−

Now, three rational numbers

equivalent to 13−

are 26−

; 39−

and

412−

.

(ii) 2 5

−−

= 2 2 5 2

− ×− ×

= 4 10−−

2 5

−−

= 2 3 5 3

− ×− ×

= 6 15−−

2 5

−−

= 2 4 5 4

− ×− ×

= 8

20−−

Now, three rational numbers

equivalent to 2 5

−− are

4 10−− ,

6 15−−

and 8

20−− .

4.4.4.4.4. LCM (2, 3, 4, 8) = 24

12−

= 1 122 12− ×

×=

1224−

14−

= 1 64 6− ×

× =

624−

23 =

2 83 8××

= 1624

58−

= 5 38 3− ×

×=

1524−

Since 16, – 6, – 12, – 15 are indescending order.

Therefore, 1624 ,

624−

, 1224−

, 1524−

are

in descending order.


Therefore, 23 ,

14

−,

12

−,

58

− are in

descending order.

5.

6. (i)2512

+ (– 1) = 2512

– 1 = 2512

– 11

= 25 12

12−

= 1312

= 11

12.

(ii) 35

− +

37

+ 6

35−

= 35

− +

37

– 635

= 21 15 6

35− + −

= 1235

−.

7. (i) 1230

− –

715

= 12 14

30− −

= 26

30−

= 13

15−

.

(ii) 1

26−

– ( )4 13−

= 1

26−

+ 413

= 1 826

− + =

726 .

8. (i) 6

13−

× 26 12

−−

= 6

13−

× 2612

= 6

12−

× 2613

= 12

− ×

21

= 22

−

= – 1.

(ii) 1519

− ÷

3038

= 1519

− ×

3830

= 1530

− ×

3819

= 12

− ×

21

= 22

− = – 1.

9. A non-zero number and its reciprocalare multiplicative inverse each other.

So, the required fraction

= Reciprocal of 2 5

−−

= 5 2

−−

.

10. (i)2

10 + ( ) 12

15−

+ ( ) 920−

= 15

+ ( ) 45

− + ( ) 9

20−

= 4 ( 16) ( 9)

20+ − + −

= 4 16 9

20− −

= 4 25

20−

= 2120

− or – 1

120

.

(ii) 217 + ( ) 3

14−

+ ( ) 128−

+ 114

= 157

+ ( ) 314−

+ ( ) 128−

+54

= 60 ( 6) ( 1) 35

28+ − + − +

= 95 7

28−

= 8828

= 227

= 317

.


WORKSHEET –73

1.1.1.1.1. Three rational numbers are

611−

× 22

, 6

11−

× 33

and 6

11−

× 44

i.e., 1222

−,

1833

− and

2444

−.

2.2.2.2.2. (i) 8 17−

= 8 ( 1) 17 ( 1)

× −− × −

(Multiplying numerator and denominator by – 1)

= 8

17−

.

(ii) 21 25−

= 21 ( 1) 25 ( 1)

× −− × −

(Multiplying numerator and denominator by – 1)

= 2125

−.

3.3.3.3.3. 1516

−=

1516

− ×

6 6

−−

(∵ – 96 ÷ 16 = – 6)

= 15 6 (16 6)

×− ×

= 90 96−

.

4.4.4.4.4.19 5−

= 19 5−

× 2 2

−−

(∵ – 38 ÷ 19 = – 2)

= (19 2)5 2

− ×× =

3810

−.

5.5.5.5.5. Since – 3 < – 2712 < – 2

Therefore, – 2712 is situated between

– 3 and – 2 on a number line.

6 .6 .6 .6 .6 . True. As negative numbers andpositive numbers are on opposite sides

of zero on the number line, 1

16 and – 1

are on opposite sides of zero on thenumber line.

7.7.7.7.7. (i) 3 14−

= 3

14−

× 2121

= 63

294−

521−

= 5

21−

× 1414

= 70

294−

∵ – 63 > – 70 ∴ 63

294−

> 70

294−

∴ 3 14− >

521−

i.e., 3 14− is greater

(ii) 59

−=

59

− ×

1616

= 80

144−

11 16−

= 11 16−

× 9 9

−−

= 99

144−

∵ – 80 > – 99 ∴ 80

144−

> 99

144−

∴ 59

− >

11 16−

i.e., 59

− is greater.

8.8.8.8.8. Absolute value of 53

− is

53 .

9.9.9.9.9.79 is the absolute value of

79

− and

79

itself.

10.10.10.10.10. (i) Let us first find LCM of 4, 8, and 12.2 4, 8, 122 2, 4, 62 1, 2, 33 1, 1, 3

1, 1, 1


∴ LCM = 2 × 2 × 2 × 3 = 24

Now, 34

− +

78

+ 1112

−

= 18 21 22

24− + −

= 1924

−.

(ii) Let us first find LCM of 3, 5 and 15.

∴ LCM = 3 × 5 = 15

Now, 8

15 +

35

− +

13

−

= 8 9 5

15− −

= 6

15−

= 25

−.

11. (i) 410

× 5

12−

× 25

= 4

12 ×

55

− ×

210

= 13

× ( 1)

1−

× 15

= 1 ( 1) 1

3 1 5× − ×

× ×

= 1

15−

(ii) 6

11−

÷ 2422

− =

611−

× 22 24−

= 611

× 2224

= 624

× 2211

= 14

× 21

= 24

= 12

.

(iii) Let us first find LCM of 8, 12 and 9.

∴ LCM = 2 × 2 × 2 × 3 × 3 = 72

Now, 18

+ 5

12 + ( ) 29

−

= 9 30 16

72+ −

= 39 16

72−

= 2372

.

(iv) 315

× 5

11 × 1

16

= 165

× 5

11 × 76

= 16 711 6

××

= 8 711 3

××

= 5633 = 1

2333 .

12. Let the reciprocal of 7 13

−− is x.

Then 7 13

−−

× x = 1 or 713

x =

11

∴ x = –13– 7 .

WORKSHEET –74

1. 2 is the absolute value of – 21 and

21

itself.

2. Absolute values less than 4 can be 3, 2and 1. Therefore, all required rationalnumbers are 3, – 3, 2, – 2, 1 and – 1.

3. – 3 = – 3 × 33

= 93

−

– 4 = – 4 × 33

= 123

−

123

− <

113

− <

103

− <

93

−.

So two rational numbers are 113

− and

103

− (Answer may very).

3 3, 5, 155 1, 5, 5

1, 1, 1

2 8, 9, 122 4, 9, 62 2, 9, 33 1, 9, 33 1, 3, 1

1, 1, 1


4. (i) 1220

− = 3 45 4

− ×× =

35

−

(ii) 1015

−=

2 53 5

− ×× =

23

−

(iii) 4480

−=

11 420 4

− ×× =

1120

−.

5. (i)

(ii)

6. (i)∵ 17 < 71

∴ – 17 > – 71.

(ii) 63

− =

6 23 2

− ××

= 126

−

∵ 3 < 12

∴ – 3 > – 12

∴ 36

− >

126

− or

36

− >

63

−.

7. (i)1118 × (– 9) +

14

− ×

56

= – 11 9

18×

– 5

4 6×

= – 112

– 5

24 =

– 11 12 – 5 124

× ×

= – 137

24 = – 5

1724 .

(ii) ( )21 39 7

× – ( )7 168 14

×

= ( )21 37 9

× – ( )7 1614 8

×

= ( )13

3× – ( )1

22

×

= 1 – 1 = 0.

(iii) ( ) 3 42 5

− × + ( )9 105 3

−× – ( )1 32 4

×

= ( ) 3 45 2

− × + ( ) 10 95 3

− × – ( )32 4×

= ( ) 32

5− × + (– 2 × 3) –

38

= 65

− –

61

– 38

= 6 8 6 40 3 5

40− × − × − ×

= 48 240 15

40− − −

= 30340

− = – 7

2340 .

8. (i)12

– 34

= 24

– 34

= 14

−

Reciprocal of 14

− =

4 1−

.

(ii)58

× 3

10−

= 5

10 ×

38

−=

12

× 38

−

= 3

16−

.

Reciprocal of 3

16−

= 16 3−

= 163

−

= – 513

.


WORKSHEET –75

1. (i) Reciprocal of 8

13−

= ( )1 8

13− =

13 8−

= 138

−.

(ii) Reciprocal of 3 5

−− = ( )

1 3 5

−−

= 5 3

−−

= 53 .

2. (i)1528 ×

1199

−=

159

× 11928

−

= 53

× 174

−

= 5 173 4

− ××

= 8512

−.

(ii) 1920

− ×

30 57

−−

= 1920

− ×

3057

= 1957

− ×

3020

= 13

− ×

32 =

12

−.

(iii) 393

− ×

145 ×

1256

−

= 393 ×

145 ×

1256

= 131

× 14

× 125

= 13 × 35

= 395

.

3. (i) 7

15−

× 5 28−

= 715

× 5

28 =

728 ×

515

= 14

× 13

= 1

12.

(ii) 5512

− ×

9633

−=

5512

× 9633

= 5533

× 9612

= 53

× 81

= 403

= 1313 .

4. Let x would be added, then

x + 4

15−

= 58

−

∴ x = 4

15 +

58

−

= 4 8

15 8××

+ 5 158 15

− ××

= 32

120 +

75120−

= 32 75

120−

= 43

120−

.

5. Let x would be subtracted. Then

45

– x = 23

∴ x = 45

– 23

= 4 35 3

××

– 2 53 5

××

= 1215

– 1015

= 2

15.

6. Let y would be added. Then

y + 78

+ 45

= – 715

∴ y = – 78

– 45

– 715

= – ( )7 4 78 5 15

+ + = – 105 96 56

120+ +

= – 257120

= – 217120

.


7. Let x should be added. Then

x + 12

+ 13

+ 15

= 8

or x + 15 10 6

30+ +

= 8

[∵ LCM (2, 3, 5) = 30]

or x + 3130

= 8

∴ x = 8 – 3130

= 240 31

30−

= 20930

or x = 62930 .

8. 34

− –

1 8−

+ 1112

– 1

16−

+ 0 – 1

16−

−

= 34

− +

18

+ 1112

+ 1

16 –

116

= 34

− +

18

+ 1112

= 18 3 22

24− + +

= 724

.

9. (i) ( )25 24 5

× – ( ) 1 105 3

− −×

= ( )25 25 4

× – ( )1 103 5

×

= ( )15

2× – ( )1

23

×

= 52

– 23

= 15 4

6−

= 116

= 156 .

(ii) ( ) 5 729 200

− ×− – ( )11 36

18 77×

+ ( )18 52 13 21

−×−

= ( )72 59 200

× – ( )11 3677 18

×

+ ( )52 1813 21

×

= ( )18

40× – ( )1

27

× + ( )64

7×

= 15

– 27

+ 247

= 7 10 120

35− +

= 11735 = 3

1235 .

10. Distance = Time × Speed

= 8 × 419

= 8 × 379

= 296

9

= 3289 km.

WORKSHEET –76

1. Let by x would be multiplied. Then

x × 1621

−=

47

∴ x = 47

× 21 16− =

4 16− ×

217

= 1 4−

× 3 = 34

−.

2. (i) 1 ÷ 18 = 1 ×

81 =

1 81×

= 8.

(ii) 6 ÷ 49

−= 6 ×

9 4−

= 6 4− ×

91

= 3 2−

× 9 = 272

− = – 13

12 .

(iii) 8

15−

÷ 163

−=

815−

× 3 16−

= 8 16

−−

× 3

15

= 12

× 15

= 1

10.


3. Let the required number be x. Then

– 12 × x = 84

∴ x = 84 12−

= 7 1−

= – 7.

4. Let the required number be y. Then

712 × y =

5018

−

Multiplying both sides by 127 , we get.

127

× 712

× y = 127

× 5018

−

or y = 2 ( 50)

7 3× −

× = 10021

−

= –41621 .

5. Let the required number be x. Then

811−

× x = 1255

−

or x = 1255

− ×

11 8−

= 1155

× 128

= 15

× 32

= 3

10.

6. Let the required number be y. Then

1845

− × y = 90 or

25

− × y = 90

∴ y = 90 × 5 2−

= – 45 × 5

= – 225.

7. Sum of 97

− and

1514 is

S = 97

− +

1514

= 18 15

14− +

= 3

14−

Product of 97

− and

1514 is

P = 97

−×

1514 =

9 157 14

− ××

Now, S ÷ P = 3

14−

÷ 9 157 14

− ××

= 3

14−

× 7 14 9 15

×− ×

= 3 9

−− ×

715 ×

1414

= 13 ×

715 × 1 =

745

Thus, SumProduct

= 745

.

8. Sum = 125

− +

1815

− =

125

− +

65

−

= 12 6

5− −

= 185

−

Difference = 125

− – ( ) 18

15−

= 1815 –

125 =

18 3615−

= 1815

−

Now,Sum

Difference=

185 1815

−

−

= 185

− ×

15 18−

= 3.

9. Let x would be added. Then

x + ( )1 1 12 3 4

+ + = 10


or x + ( )6 4 312

+ += 10 or x +

1312 = 10

or x = 10 – 1312

= 120 13

12−

= 10712

= 81112 .

10. Let the other number be y. Then

159

− + y = – 10 or

53

− + y = – 10

∴ y = – 10 + 53

= 30 5

3− +

= 253

−.

11.15

− = 15

Reciprocal of15

− = Reciprocal of 15

= 51

= 5.

❏❏


WORKSHEET –77

1. (C) A regular polygon has as many

lines of symmetry as it has sides.

2. (A) A circle has infinitely many

number of lines of symmetry. Each of

them passes through its centre.

3. (D) A square has a rotational symmetry

of order 4 about its centre.

4. (C) The letter ‘H’ has the reflectional

symmetry about both the horizontal

and vertical mirrors as shown below.

5. (D) Here, 360° is divisible only by 24°.

6. (C) A parallelogram has no line ofsymmetry.

7. (A) A rhombus has 2 lines of symmetryand rotational symmetry of order 2.

8. (D) A circle has rotational symmetryof infinite order.

9. (A) The least angle = 360

5°

= 72°.

10. (D) Since BC + CA < AB, so the triangleis not possible.

11. (A) Each angle of an equilateral triangleis of measure 60°.

12. (B) Given BC = 5 cm.

13. (D) ∵ DA y BC and AB is transversal.

∴ ∠ DAB = ∠ ABC


14. (C) We can join C to any point on AB.

15. (A) Each angle of an equilateral triangle

is of measure 60°.

16. (B) To construct any triangle, we firstdraw a side.

17. (A) We should first draw BC = 6 cmbecause the given angle is on one endof BC.

WORKSHEET –78

1. Line m is the required line.

11Chapter

SYMMETRY AND PRACTICAL GEOMETRY

129S Y M M E T R Y A N D P R A C T I C A L G E O…

2. Line AD passes through A such thatAD y BC.

3. Here is the rough sketch of the triangleXYZ.

Construction:

Step 1. Draw XY = 3 cm.

Step 2. Taking X as centre and radiusof 3 cm, draw an arc.

Step 3. Taking Y as centre and radiusof 4 cm, draw another arc.

Step 4. The arcs obtained in step 2.and step 3, intersect each other at Z.

Step 5. Join XZ and YZ.

XYZ is the required triangle.

4. Here is a rough sketchof the triangle PQR.

Construction:

Step 1. Draw a linesegment PQ = 2.5 cm

Step 2. Make an angleof measure 110° at Q such that ∠ PQX= 110°.

Step 3. Taking Q as centre and radiusof 4.5 cm, draw an arc to cut QX at R.

Step 4. Join PR

PQR is the required triangle.

5. Here is a rough sketch of the triangle.

Constructions:

Step 1. Draw a line segment AB = 4 cm.

Step 2. Make an angle of measure 90°at the end A such that ∠BAX = 90°.


Step 3. Taking B as centre and radiusof 6 cm, draw an arc to intersect theray AX at C.

Step 4. Join BC.

ABC is the required triangle.

6. Here is a rough sketch of ∆ PQR.

Yes. The triangle is possible by ASA

criterion.

7. (i) A circle has infinitely many axes of

symmetry passing through its

centre.

Here, we are drawing an axis of

symmetry namely XY, of the given

circle having centre at O.

(ii) A parallelogram has no axis ofsymmetry.

WORKSHEET –79

1. (i)

There are six lines of symmetry,namely A1A2, B1B2, C1C2, D1D2, E1E2

and F1F2.

(ii) There are two lines of symmetry,namely P1P2 and Q1Q2.


2.

3. (i) An isosceles triangle has one line ofsymmetry.

(ii) A regular hexagon has six lines ofsymmetry.

4. (i) The figure has rotational symmetryof order 6.

(ii) The figure has rotational symmetryof order 4.

5. The two examples are: (a) a parallel-ogram and (b) a scalene triangle.

6. A parallelogram has a rotationalsymmetry of order 2 but no line ofsymmetry.

7. Yes.Sum of two sides = 10 cm + 8 cm

= 18 cm which is greater than thirdside.

So, the triangle is possible.

8. Construction:

Step 1. Draw a horizontal line l.

Step 2. Mark two points A and B on l.

Step 3. Draw two perpendiculars AMand BN on the line l.

Step 4 Mark two points C and D onrespectively AM and BN such that AC= BD = 3.5 cm.

Step 5. Join CD and extend it to bothsides, call it line m.

The lines l and m are required linessuch that l y m.



10. (i) A circle has infinitely many num-ber of lines of symmetry.

(ii) A rectangle has two lines of symm-etry.

11. AB is given line segment and m is itsline of symmetry.

12. Centre of Rotation:

A fixed point about which an objectrotates is called the centre of rotation.

WORKSHEET – 80

1. Angle of Rotation:

The least angle through which rotatingan object about a fixed point, it appearsin the same position is called the angleof rotation.

2. (i) It is a scalene triangle, so there is noline of symmetry.

(ii) There is one line of symmetry whichis shown as dotted line.

(iii) There is one line ofsymmetry, which is shownas dotted line.

3. The measures of all the three sides ofan equilateral triangle are equal.

Here is a rough sketch of the triangle.

4.

Here, l y m y n.

5. One.

6. (i) Line l is the required line ofsymmetry.


(ii)

Lines m and n are the two requiredlines of symmetry.

7. The order of rotational symmetry of aregular octagon is 8.

8. Figure (ii) is a square which has morethan one i.e., four lines of symmetry.

9. (i)

(ii)

10. A regular pentagon has 5 lines ofsymmetry, namely A1A2, B1B2, C1C2,D1D2 and E1E2.


WORKSHEET – 81

1.

Object Centre of Order of Angle of

of rotation rotation rotation(i) circle Centre of Infinitely Slightly

the circle many greaterthan zero

(ii) Rhombus point ofintersection 2 180°

of diagonals



Construction:

Step 1. Draw a line segment AB= 7 cm.

Step 2. Make an angle of 90° at A suchthat ∠BAX = 90°.

Step 3. Taking A as centre and radiusof 5 cm, draw an arc to cut AX at C.

Step 4. Join BC.

∆ ABC is the required triangle.

3. XY + YZ = 3 cm + 4 cm = 7 cm

Yes, ∆XYZ is possible as

XY + YZ > XZ.

4. ABCD is theoriginal squareand AB’ C’ D’ isthe requiredsquare.


Construction:

Step 1. Draw a line segment QR= 3 cm.

Step 2. Make an angle of 90° at Q suchthat ∠RQX = 90°.

Step 3. Taking R as centre and radiusof 5 cm, draw an arc to intersect theray QX at P.

Step 4. Join PR.

PQR is the required triangle.

6. (i) A circle has infinitely many lines ofsymmetry.


(ii) A regular hexagon has 6 lines ofsymmetry.

7. (i) Order = 1 (ii) Order = 3.

8. The line segment AB and its line of

symmetry XY is shown here.

9. Order = 3.

10. (i) Lines of symmetry are: A1A2, B1B2,C1C2 and D1D2.

(ii) Line of symmetry is XY.

11. Line l is parallel to the given line m,i.e. l y m.

WORKSHEET –82

1. Here is a rough sketch of the ∆ABC.


Construction:

2.

.

3. (i) Number of lines of symmetry = 1.(ii) Number of lines of symmetry = 1.

4. No. According to the angle sumproperty of a triangle, the total measureof the internal angles is 180°. But, herethe sum of only two angles is morethan 180° as ∠E + ∠F = 90° + 110° =200°. So, the construction is notpossible.

5.

For ∆∆∆∆∆ABC: AB = 5.7 cm, BC = 4.5 cm,CA = 5.3 cm.

Here, BC + CA = 4.5 cm + 5.3 cm

= 9.8 cm.

So, BC + CA > AB.

For ∆∆∆∆∆PQR: PQ = 10 cm, QR = 6.6 cm,RP = 5.7 cm

Here, QR + RP = 6.6 cm + 5.7 cm

= 12.3 cm.

So, QR + RP > PQ

Thus, in each case the sum of two sidesis greater than the third side.



Construction:


Construction:

8. Shapes (i) and (ii) are the examples ofreflectional symmetry.

Shape (ii) is the example of rotationalsymmetry.

9. (a) Rectangle (b) rhombus.

10. Here is a rough sketch of triangle

Construction:

11. (i) Sum of measures of two sides = 12 cm + 1 cm = 13 cm.

Measure of third side = 14 cm

∵ Sum of measures of two sides <Measure of third side

So, triangle is not possible.

(ii) Sum of measures of two sides = 1 m + 3 m = 4 m.


Measure of third side = 8 m∵ Sum of measures of two sides <Measure of third side

So, triangle is not possible.

12. Order = 6.

WORKSHEET –83

1. (i) One line of symmetry

(ii) Four lines of symmetry

(iii) One line of symmetry

(iv) One line of symmetry

2. (i) One line of symmetry

One line of symmetry

3. (i) Order = 4.

(ii) It is a regular pentagon∴ Order = 5.

4. Shapes (i), (ii), (iv), (v) and (viii) haverotational symmetry.

5. (i) Order = 3.

(ii) Order = 1.6. You are given a line CD and two points

P and Q on either sides of it.Construction:

Step 1. Join PC

Step 2. Taking C as centre and anyconvenient radius, draw an arc tointersect CD at M and CP at N.

Step 3. Taking P as centre and sameradius as in step 2, draw an arc RS tointersect CP at R.Step 4. Place the pointed tip of thecompasses at M and adjust the openingso that the pencil tip is at N.Step 5. Taking R as centre and openingsame as in step 4, draw an arc to cutthe arc RS at T.Step 6. Join PT and extend it to bothsides.Step 7. Repeat the process from step 1to step 6 for the point Q. You wouldfind a line QU.Thus, PT y CD and QU y CD.


7. No, the triangle is not possible .Reason:∠B + ∠C = 90° + 95° = 185°.According to the angle sum propertyof a triangle, “total measure of theinterior angles of a triangle is 180°.”Therefore, for the given data, angle sumproperty of the triangle does not hold.

8. (i) Sum of two sides = 8 cm + 6 cm= 14 cm

Since, the sum of two sides is notgreater than third side.Therefore, the triangle is notpossible.

(ii) Sum of two sides = 2 cm + 4 cm

= 6 cm

Since the sum of two sides is greaterthan third side. Therefore, thetriangle is possible.

9. Shapes (i) and (ii) both are examplesof reflectional symmetry.

Shape (i) is the example of rotationalsymmetry.

10. H, I, O and X

11. Other angles will be 120°, 180°, 240°,300° and 360°.

❏❏


WORKSHEET –84

1. (A) Perimeter = 4 × Side

= 4 × 4 = 16 cm.

2. (B) l = 25 cm; b = 6 cm

Perimeter = 2 × (l + b) = 2 (25 + 6)

= 2 × 31 = 62 cm.

3. (A) Area = Side × Side

= 2.1 × 2.1= 4.41 cm2.

4. (D) Length of wire

= Circumference of the pipe

= 2p × Radius

= 2 × 272 × 7 = 44 cm.

5. (A) We know that the value of p isabout 3.141592.

So, the approximate value of p is 3.14.

6. (C) 1 m2 = 1 m × 1 m

= 100 cm × 100 cm

= 10000 cm2.

7. (A) Area of parallelogram

= Base × Height

= 6 × 2.2

= 13.2 cm2.

8. (D) Base = 60 cm = 60

100 m =

610

m.

Height = 80 cm = 80

100 m =

810

m.

Area of ∆PQR = 12

× Base × Height

12Chapter

PERIMETER AND AREA

= 12

× 6

10 × 8

10

= 24

100 = 0.24 m2.

9. (A) r = 2d

= 15.4

2= 7.7 cm

Circumference = 2pr = 2 × 227 × 7.7

= 48.4 cm.

10. (B)Area of the shaded region

= Area of outer circle – Areaof inner circle

= p (8)2 – p (4)2

= 64 p – 16 p = 48 p= 48 × 3.14 = 150.72 m2.

Cost of polishing = 150.72 × 3

= ` 452.16.

11. (B) AB = DC = 4 cm

Area of ∆ ABD = 12

× Base × Height

= 12

× AB × MD

= 12

× 4 × 6 = 12 cm2.

12. (C) Base = 2 cm = 2 × 10 mm.

Height = 1.1 cm = 1.1 × 10 mm

= 11 mm.

Area = 12

× Base × Height

= 12

× 2 × 10 × 11 = 110 mm2.

141I EM ET RIREP EM ET R A DN A AER

13. (A) AB = BC = 6 mm

Area of the ∆ABC = 12 × AB × CD

∴ CD = 2 Area of ABC

AB× ∆

= 2 13.2

6×

= 4.4 mm.

14. (C)Area of the ∆PQR = 12

× PQ × OR

= 12

× 3 × 2

= 3 cm2.

15. (A) Area of a parallelogram

= Base × Height.

16. (B) Perimeter of rectangular sheet = Perimeter of the squared sheetor 2(length + breadth) = 4 × side

∴ 2(60 + breadth) = 4 × 40

∴ Breadth = 160

2 – 60

= 20 cm.

Area of the rectangular sheet

= Length × Breadth

= 60 × 20 = 1200 cm2.

17. (D) Length = Area

Breadth =

284

= 7 cm.

WORKSHEET –85

1. Perimeter of a square = 4 × Side

∴ 440 = 4 × Side

∴ Side = 4404

= 110 m.

Area of a square = Side × Side

= 110 × 110

= 12100 m2.

2. Side = 60 m

(i) Area = Side × Side = 60 × 60

= 3600 m2.

(ii) Since the wire is fenced four times,

∴ Length of the wire

= 4 × Perimeter of land

= 4 × (4 × Side)

= 16 × Side

= 16 × 60 = 960 m.

Cost of fencing

= Length × Rate of 1 m of wire

= 960 × 25 = 24000

Therefore, the cost of fencing is` 24, 000.

(iii) Total cost of the land

= Area of the land × Cost of 1 m2

= 3600 × 10500 [Using part (i)]= 36 × 105 × 10000

= 3780 × 10000

= 37800000.

Therefore, the total cost of the land is` 3,78,00,000.

3. Perimeter = 520 m, breadth = 40 m.

We have, perimeter of a rectangle

= 2(length + breadth)

∴ 520 = 2 × (length + 40)

∴ Length = 260 – 40 = 220 m

And area = Length × Breadth

= 220 × 40 = 8800 m2.

4. Area = 84.8 cm2, base = 4 cm

Area of a parallelogram

= Base × Height

∴ 84.8 = 4 × height

or height = 84.8

4 = 21.2 cm.


5. r1 = 3.5 cm, r2 = 7 cm

(i) Diameter, d1 = 2 × r1 = 2 × 3.5

= 7 cm

Diameter d2 = 2 × r2

= 2 × 7 = 14 cm.

(ii) Circumference,

C1 = 2pr1 = 2 × 227 × 3.5

= 2 × 22 × 0.5 = 22 cm

Circumference,

C2 = 2p r2 = 2 × 227 × 7

= 2 × 22 = 44 cm.

(iii) Ratio of circumferences

= 1

2

CC

= 2244

= 12

= 1 : 2.

6. Length of the outer rectangle

= 100 m + 5 m + 5 m

= 110 m.

Breadth of the outer rectangle

= 80 m + 5 m + 5 m

= 90 m.

Area of the path = Area of the shadedregion = Area of the outer rectangle –Area of the inner rectangle

= 110 × 90 – 100 × 80

= 9900 – 8000

= 1900 m2.

7. Radius of circle, r = 28 cm

The straight edge of the shaded part isthe diameter of the circle, whichdivides the circle into two halves.

∴ Area of the shaded part

= 12

× Area of the circle

= 12

× pr2

= 12

× 227

× 28 × 28

= 11 × 4 × 28

= 1232 cm2.

Thus, the area of the shaded part is1232 cm2.

8. Let the required number of discs be n.

Since the thicknesses of both types ofthe sheets are same, therefore, theirareas must be equal.

∴ Area of n discs

= Area of the rectangular sheet

or n × p × (radius)2 = Length × Breadth

or n × 227

× 14 × 14 = 56 × 33

or n = 56 33

22 2 14×

× ×

or n = 5628

× 3322

or n = 2 × 32

= 3

Thus, the required number of discs is 3.


WORKSHEET –86

1. Square. If we increase the perimeter ofa square, then its side increases and soits area increases.

2. (i) l = 15 cm, b = 5 cmPerimeter = 2 ×(l + b) = 2 × (15 + 5)

= 2 × 20 = 40 cm.Thus, perimeter of the rectangle is40 cm.

(ii) l = 8 cm, b = 2.2 cmPerimeter = 2 × (l + b) = 2 × (8 + 2.2)

= 2 × 10.2 = 20.4 cm.Thus, perimeter of the rectangle is20.4 cm.

3. (i) Perimeter of a square = 4 × Side= 4 × 15= 60 cm.

Thus, perimeter of the square is60 cm.

(ii) Perimeter of a square = 4 × Side= 4 × 0.9

= 4 × 9

10

= 3610 = 3.6 cm.

Thus, perimeter of the square is3.6 cm.

4. Area of a square = Side × Side∴ 49 = Side × Sideor 7 × 7 = Side × Side∴ Side = 7 cm.

5. Breadth = 4.2 cmBy question, length = 2 × Breadth

= 2 × 4.2 = 8.4 cmPerimeter = 2 × (l + b)

= 2 × (8.4 + 4.2)= 2 × 12.6= 25.2 cm.

Thus, perimeter of the rectangle is25.2 cm.

6. Side of the square field = 12.5 m

Perimeter of the field = 4 × Side

= 4 × 12.5

= 5 m.

Since Romi runs 3 times around thesquare field.

∵ Distance covered by Romi

= 3 × Perimeter

= 3 × 50 = 150 m

Thus, the distance covered by Romi is150 metres.

7. Let b = x (say)

Then, l = 3 × b = 3x

Here, perimeter = 2 × (l + b)

Given Perimeter = 64 m

∴ = 2(3x + x) = 64

or = 8x = 64

∴ x = 648 = 8

∴ 3x = 3 × 8 = 24.

Thus, the length of the room is 24 mand the breadth is 8 m.

8. In ∆ABC, ∠C = 90°

∴ AB2 = BC2 + CA2


= 32 + 42 = 9 + 16 = 25 = 5 × 5


∴ AB = 5 cm

Now, perimeter = AB + BP + PQ + AQ

= 5 + 6 + 5 + 6

= 22 cm.

Thus, the perimeter of the figure is22 cm.

9. (i) Area of ∆ABC

= 12 × BC × AB

= 12 × 3 × 4

= 6 cm2.

(ii)

Area of ∆ABC = 12 × AC × AB

= 12 × 4 × 2 = 4 cm2.

10.

Area of parallelogram ABCD

= Base × Height

= AB × CE

= 7.2 × 4.5

= 72 45

100×

= 3240100

= 32.40 cm2.

11. Let b = x, then, l = 3 × b = 3 × x = 3x

Perimeter = 2 × (l + b)

∴ 26.4 = 2 × (3x + x)

or26410

= 2 × 4x

∴ 26410 8×

= x

or x = 3310

= 3.3

∴ 3x = 3 × 3.3 = 9.9

So, the length of the room is 9.9 m andthe breadth is 3.3 m.

WORKSHEET –87

1. Perimeter of square = 4 × Side

= 4 × 42.5 = 170 m

Distance covered by Saloni

= 8 × Perimeter of the field

= 8 × 170

= 1360 m.

2. Let breadth = x, then length = 3x

Perimeter = 2 × (length + breadth)

∴ 164 = 2 × (3x + x)

or824

= x

or x = 20.5

∴ 3x = 3 × 20.5 = 61.5.

Therefore, length of the hall is 61.5 mand the breadth is 20.5 m.

3. (i) r = 2d

= 1.42

cm

C = 2pr = 2 × 3.14 × 1.42

= 3.14 × 1.4

= 4.396.

Thus, the circumference is 4.396 cm.


(ii) r = 2d

=292

C = 2pr = 2 × 3.14 × 292

= 3.14 × 29 = 91.06

Thus, the circumference is 91.06 mm.

4. (i) C = pd.

⇒ 4.2 = 3.14 × d

⇒ d = 4.2

3.14 =

420314

⇒ d = 1.337

⇒ d = 1.34 cm.

(ii) C = pd

⇒ 252 = 3.14 × d

⇒ d = 2523.14

= 25200

314= 80.254 = 80.25 mm.

5. (i) C = 2pr

⇒ 77 = 2 × 3.14 × r

⇒7700628 = r

∴ r = 12.261

∴ r = 12.26 cm.

(ii) C = 2pr

⇒ 126 = 2 × 3.14 × r

∴ 12600628

= r

∴ r = 20.063

∴ r = 20.06 mm.

6. r = 40 cm

Area = p r2 = 3.14 × 40 × 40

= 314 × 16 = 5024 cm2.

7. R = 14 m, r = 2.8 mm

Area of washer

= Area of the outer circle –Area of the inner circle

= p R2 – p r2

= p (R2 – r2)

= 227 × (142 – 2.82)

= 227

× 188.16 = 591.36 mm2.

8. (i) The figure contains one rectangleand two semicircles of diameter7 cm.

Perimeter = Curve ABC + CD +Curve DEF + FA

= p r + CD + p r + CD

[∵ CD = FA]

= 2p r + 2 CD

= 2 × 227 ×

72 + 2 × 10

= 22 + 20 = 42 cm.


(ii) The figure contains a square and 4semicircles of diameter 14 cm each.

Perimeter = 4 × length of curvedpart of one semicircle

= 4 × p × 2d

= 4 × 227 ×

142 = 88 cm.

9. Area of circle = p r2 = 227

× (7.7)2

= 227

× 7.7 × 7.7

= 186.34 cm2.

Area of square = Side × Side = 7 × 7

= 49 cm2.

Therefore, the circle has more area.

10. Perimeter = 4 × Side

∴ 36 = 4 × Side

∴ Side = 364

= 9 cm.

Area of square = Side × Side = 9 × 9

= 81 cm2.

WORKSHEET –88

1.

AF = BC + DE

= 1 + 2.5 = 3.5 cm

Perimeter = AB + BC + CD + DE + EF+ AF

= 7 + 1 + 4 + 2.5 + 3 + 3.5

= 21 cm.

2. Area = AB × BC

= 11.2 × 2.5 = 28 cm2.

3. (i) Area = a2 = (12.6)2 = 12.6 × 12.6

= 158.76 cm2

(ii) Area = a2 = 42 = 4 × 4 = 16 cm2.

4. (i)

Area = 12 × Base × Height

= 12 × BC × AB

= 12 × 8 × 3 = 12 cm2.

(ii) In ∆ABC,

∠A = ∠B

∴ BC = AC = 7 cm

Further, ∠A + ∠B + ∠C = 180°

∴ ∠C = 180° – 45° – 45°

= 90°

So ∆ABC is a right-angled triangle.

∴ Area of ∆ABC

= 12 × Base × Height


= 12 × AC × BC

= 12 × 7 × 7 =

492

= 24.5 cm2.

5. Base = AB = 12.5 cm

Height = BD = h (say)

Area = AB × h

∴ 500 = 12.5 × h

∴ h = 50012 5. =

5000125 = 40 cm.

Thus, height of the parallelogram is40 cm.

6. Length of the inner rectangle

= 100 m – 3 m – 3 m

= 94 m.

Breadth of the inner rectangle= 90 m – 3 m – 3 m= 84 m.

Area of the path= Area of the shaded region= Area of the outer rectangle

– Area of the inner rectangle

= 100 × 90 – 94 × 84

= 9000 – 7896 = 1104 m2.

Thus, area of the path is 1104 m2.

7. Side of inner square

= 200 m – 10 m – 10 m = 180 m

Area of the path

= Area of the shaded region

= Area of the outer square

– Area of the inner square

= 2002 – 1802

= 200 × 200 – 180 × 180

= 40000 – 32400

= 7600 m2.

8. Breadth = 0.25 m = 0.25 × 100 cm

= 25 cm

Area of a rectangle = Length × Breadth

∴ 2500 = Length × 25

∴ Length = 2500

25 = 100 cm.

Perimeter of the sheet

= 2 × (length × breadth)

= 2 × (100 + 25)

= 2 × 125 = 250 cm.

9. Length of fencing

= 20.8 + 12.5 + 12.5

= 45.8 m.


Cost of fencing

= Length × Rate per metre.

= 45.8 × 125 = 458 125

10×

= 57250

10 = ` 5725

Thus, the cost of fencing is ` 5725.

10. (i) Radius of the circle, r1 = 35 cm. Theshaded part of the circle is itsquadrant.

∴ Area of the shaded region

= 14 × pr1

2

= 14 ×

227 × 35 × 35

= 112 × 5 × 35

= 1925

2 = 962.5 cm2.

(ii) Radius of the circle, r2 = 7 cm

The shaded part of the circle is itsquadrant.

∴ Area of the shaded part

= 14 × p r2

2

= 14 ×

227 × 7 × 7

= 112 × 7 =

772 = 38.5 cm2.

11. Circumference = p d

= 3.14 × 1.8

= 5.652 cm2.

WORKSHEET –89

1. Join HC. ABCH is a rectangle

∴ HA = BC = 4 cm

GDEF is also a rectangle

∴ FG = DE = 1.2 cm

Perimeter

= AB + BC + CD + DE + EF + FG+ GH + HA

= 2 + 4 + 1.5 + 1.2 + 5 + 1.2 + 1.5 + 4

= 20.4 cm.

Area of the shape

= Area of the rectangle ABCH+ Area of the rectangleGDEF

= 4 × 2 + 5 × 1.2

= 8 + 6 = 14 cm2.

2. (i) PQ = Side = 1.5 cm

Perimeter of the square PQRS

= 4 × Side = 4 × PQ

= 4 × 1.5 = 6 cm.

Area of the square PQRS

= Side × Side = PQ × PQ

= 1.5 × 1.5 = 2.25 cm2.

(ii) PQ = Side = 12 mm

Perimeter of square PQRS

= 4 × Side = 4 × PQ

= 4 × 12 = 48 mm


Area of square PQRS

= Side × Side = PQ × PQ

= 12 × 12 = 144 mm2.

3. Area of square = side2

∴ 2.25 = Side × Side

or225100

= Side × Side

or1510

× 1510

= Side × Side

∴ Side = 1510

m.

Perimeter of square = 4 × Side

= 4 × 1510

= 6 m.

4. (i) Area of rectangle

= Length × Breadth

= 24 × 14 = 336 cm2.

(ii)Area of square= Side2 = 222

= 22 × 22 = 484 cm2.

Clearly, area of the square is greater.

Differences in areas

= Area of the square – Areaof the rectangle

= 484 – 336 = 148 cm2.

Thus, area of the square is 148 cm2

more than that of the rectangle.

5. l = 80 m, b = 35 m

(i) Area of the playground

= l × b = 80 × 35

= 2800 m2

Cost of levelling

= Area × cost of per square metre

= 2800 × 1.50 = 280 × 15

= 4200.

Thus, the cost of levelling theplayground is ` 4200.

(ii) Perimeter of the playground

= 2 × (l + b) = 2 × (80 + 35)

= 2 × 115 = 230 m.

Distance walked by a boy

= 2 × Perimeter

= 2 × 230 = 460 m.

Time taken by the boy

= Distance walkedSpeed

= 4601.5

= 460015

= 9203

= 306.67 seconds

or 5.11 minutes.

6. b = 15 cm, Area = 60 cm2, h = ?

Area = 12

bh

∴ 60 = 12

× 15 × h

∴ h = 60 2

15×

= 8

Thus, the altitude is 8 cm.

7. Area of a trapezium =

Sum of parallel sides × Distance between them

2

= (13 9) 9

2+ ×

= 22 9

2×

= 99 cm2.8. r = 2.1 m

Circumference = 2p r = 2 × 227 × 2.1

= 2 22 21

7 10× ××

= 2 22 3

10× ×

= 13210

= 13.2 cm2.


9. 1

2

rr

= 23

1

2

CC

= 1

2

22

rr

ππ

= 1

2

rr

= 23

Thus, the ratio of the circumferences is2 : 3.

10. Forming a new closed shape from anyclosed shape, the perimeter remainsunchanged.

∴ Perimeter of the square = Perimeterof the circle

or 4 × Side = 2pr

∴ Side = 24 ×

227 × 42

= 11 × 6

= 66 cm.

Thus, the side of the square is 66 cm.

11. ABCE is a rectangle.

∴ AE = BC = 5 m

And CE = AB = 17 m

Since CDE is a semicircle and CE is itsdiameter.

∴ CDE = 12

p × CE = 22

2 7× × 17

= 187

7 m.

Now, perimeter of the figure

= AB + BC + CDE + EA

= 17 + 5 + 187

7 + 5

= 27 + 187

7 = 3767 = 53.71 m.

WORKSHEET –90

1. Circumference of the pillow cover= p × Diameter= 3.14 × 1.4 = 4.396 m

∴ Length of the lace required= Circumference of the cover= 4.396 m

Cost = Length of the lace × rateper metre

= 4.396 × 20 = ` 87.92.

2. Circumference = 2pr = 2 × 227 × 1.4

= 44 × 0.2 = 8.8 cm.

3. Circumference of the pipe,

C = 2pr = 2 × 227 × 100

= 4400

7 cm.

(i) Length of the tape to wrap once

= C × 1 = 4400

7 × 1

= 628.57 cm.

(ii) Length of the tape to wrap twice

= C × 2 = 4400

7 × 2

= 8800

7 = 1257.14 cm.

4. Length of the inner rectangle

= 150 m – 3 m – 3 m

= 144 m.


Width of the inner rectangle

= 120 m – 3 m – 3 m

= 114 m.

Area of the path

= Area of the shaded region

= Area of the outer rectangle– Area of inner rectangle

= 150 × 120 – 144 × 114

= 18000 – 16416

= 1584 m2.

5. Length of fencing = 28.8 + 18.5 + 18.5

= 65.8 m

Cost of fencing = Length of fencing ×Rate per metre

= 65.8 × 125

= 8225

Thus, the cost of fencing is ` 8225.

6. r = 2d

= 492 cm

Area of semicircle

= 12

p r2 = 12

× 227

× ( )2492

= 117 ×

492 ×

492

= 114

× 7 × 49 = 3773

4= 943.25 cm2

Length of the boundary

= d + pr = 49 + 227 ×

492

= 49 + 11 × 7 = 49 + 77

= 126 cm.

7. (i) There are four quadrants in a circle

of radius ( )40 cm

2 = 20 cm, one at

each corner of a square with sidelength 40 cm.

Area of square

= Side × Side = 40 × 40

= 1600 cm2.

Area of 1 quadrant

= 14 p (radius)2

= 14 ×

227 × 20 × 20

= 2200

7 cm2.

Sum of areas of 4 quadrants

= 4 × Area of 1 quadrant

= 4 × 2200

7 = 8800

7 cm2.

Now, area of the shaded portion

= 1600 – 8800

7

= 800 ( )112

7− = 800 ×

37

= 2400

7 = 342.86 cm2.

(ii) Area of the circle with centre O

= p × (radius)2

= p × 7 × 7

= 49p cm2.

Area of the circle with centre at O’.

= p × (radius)2

= p × 1.4 × 1.4

= 1.96 p cm2.

Sum of the areas of both the circles


= 49 p + 1.96 p

= 50.96 × 227

= 160.16 cm2.

Area of the square having sides20 cm = (side)2 = 20 × 20

= 400 cm2.Now, area of the shaded portion

= 400 – 160.16= 239.84 cm2.

8. r = 14 cm

Perimeter of each semicircular disc

= Diameter + Circumference

2

= 2r + pr = (2 + p) × r

= ( )222

7+ × 14 = 28 + 44

= 72 cm.

9. Area of the lawn = l × b

∴ 1250 = 50 × b

∴ b = 1250

50 = 25 m.

Now, perimeter of the lawn

= 2 × (l + b)

= 2 × (50 + 25)

= 2 × 75 = 150 m.

10. Base = 2.5 cm, area = 100 cm2.

Area of a parallelogram

= Base × Height

∴ 100 = 2.5 × Height

∴ Height = 1002 5.

= 1000

25= 40 cm.

Thus, the height of the parallelogramis 40 cm.

11. Perimeter = 8 cm + 6 cm + 3 cm

= 17 cm.

WORKSHEET – 91

1. Area of a circle = pr2

⇒ 12474 = 227 × r2

⇒ r2 = 12474 7

22×

⇒ r2 = 567 × 7

= 81 × 7 × 7

or r2 = 9 × 9 × 7 × 7

∴ r = 9 × 7 = 63 cm.

2. Circumference of the circle

= Perimeter of the square

or 2pr = 4 × Side

∴ 2 × 227 × r = 4 × 11

∴ r = 4 11 7

2 22× ××

= 7 cm.

Now, area of the circle

= pr2 = 227

× 7 × 7

= 154 cm2.

3. pr2 = 15400

or227 × r2 = 15400

∴ r2 = 15400 7

22×

= 700 × 7

= 7 × 10 × 10 × 7

or r2 = 72 × 102

or r2 = (7 × 10)2

∴ r = 7 × 10 = 70 m

Circumference = 2pr = 2 × 227 × 70

= 440 m.


4. r = 142

= 7 cm

Area of the shaded portion

= Area of the rectangle – Area of the semicircle

= l × b – 12

pr2

= 14 × 9 – 12

× 227

× 7 × 7

= 126 – 77 = 49 cm2.

Thus, the area of the shaded portion is49 cm2.

5. Side of the inner square

= 150 m – 2 m – 2 m

= 146 m.

Area of the road

= Area of the outer square – Area of the inner square

= 150 × 150 – 146 × 146

= 22500 – 21316

= 1184 m2.

Cost of constructing the road

= Area of the road × Cost persquare metre

= 1184 × 3 = ` 3552

Thus, the cost of constructing the roadis ` 3552.

6. Area of the door = 132 × 200 cm2

= 26400 cm2.

= 2.64 m2

[10000 cm2 = 1 m2]

Area of the whole wall

= 250 × 200 cm2

= 50000 cm2.

= 5 m2

Area of the wall for painting

= Area of the whole wall– Area of the door

= 5 – 2.64 = 2.36 m2.

Cost of painting the wall

= 2.36 × 2.50 = ` 5.90.

7. Area of a square = Side × Side

∴ 49 = Side × Side

or 7 × 7 = Side × Side

or 72 = (Side)2

∴ Side = 7 cm.

8. (i) Area of parallelogram

= Base × Height

∴ 36 = 4 × Height

∴ Height = 364

= 9 cm.

Thus, the height of the parallelo-gram is 9 cm.

(ii) Area of parallelogram

= Base × Height

∴ 16.38 = 15.6 × Height

∴ Height = 16 3815 6..

= 16381560

= 1.05 cm

Thus, the height of the parallelogramis 1.05 cm.


9. Perimeter of square = 4 × Side

∴ 144 = 4 × Side

∴ Side = 144

4 = 36 cm

Area of square = Side × Side

= 36 × 36

= 1296 cm2.

10. Area of the outer cross-section

= p × (Radius)2

= p × 72 = 49p.

Area of the inner cross-section

= p × (Radius)2

= p × (3)2 = 9p.

Area of the cross-section of the pipe

= 49 p – 9 p

= 40 p = 40 × 3.14

= 125.60 cm2.

11. Perimeter of the figure

= 1.5 cm + 2.5 cm + 3.5 cm= 7.5 cm.

❏❏

155I MEGLA RB IA C E PX R SSE I SNO

WORKSHEET –92

1. (B) 11xy and – 6y have different algeb-

raic factors.

2. (D) The coefficient of x is – y2.

3. (B) The polynomial a2 + b2 has two

terms and hence it is a binomial.

4. (C) ( 8x2 + 4x2 ) + (– 6x – 4x + 3x) + 5

= 12x2 – 7x + 5.

5. (A) 3 74 – 57 2

xxx

++

+

6. (C) 8 – 4 18 – 6 2

– –– 10 2 –

xy x yxy x y

xy x y

++

++

7. (D) 7 – 6 8 3 – 12 910 – 18 17

pq p qpq p qpq p q

++ +

+

8. (A) 2

2

2

3 2 – 2 7

4 – 2 9

aa a

a a

++ +

+

2

2

2

– 2 34 – 2

3 2 1

a aa

a a

+ ++

+ +

Now,

13Chapter

ALGEBRAIC EXPRESSIONS

9. (A) 2

2 2

2 2

2 3 2

– – – –

x xyx xy y

x xy y

++ +

+

10. (A) p + 8 = 2 + 8 = 10.

11. (A) m – 2 = 2 – 2 = 0.

12. (D) The expression 5 is independent ofx. So, for x = 2, the expression willremain 5.

13. (C) At x = – 1,

x3 – 6x2 + 5 = (– 1)3 – 6(– 1)2 + 5

= – 1 – 6 + 5 = – 2.

14. (C) a3 – b3 = 13 – 13 = 1 – 1 = 0.

15. (A) 2x2 + x – m = 5

⇒ 2(0)2 + 0 – m = 5 (Putting x = 0)

∴ m = – 5.

16. (D) Constant term = 6.

17. (C) x3 and 7x3 has same literal factoras x3.

18. (A) a2 + b2 + ab = 22 + (– 2)2 + 2(– 2)

= 4 + 4 – 4 = 4.

19. (B) The exponent of b in a2 – b3 + 8 is 3.

20. (A) 2(x2 – x + y) + 3 (x + y)

= 2x2 – 2x + 2y + 3x + 3y

= 2x2 + x + 5y.

21. (A) (y2 – 1) is not a factor of 2z2 – 2and (z2 – 1) is not a factor of 2y2 – 2.

22. (B) Area of a rectangle= Length × Breadth= l m × b m

= lb m2.

2

2

2

4 – 2 93 2 1

– – –– 4 8

a aa a

a a

++ +

+


23. (D) P + 2Q – R

= m2 – n2 + 2(n + m) – (2m + 2n + m2)

= m2 – n2 + 2n + 2m – 2m – 2n – m2

= (m2 – m2) – n2 + (2n – 2n) + (2m – 2m)

= – n2.

WORKSHEET –93

1. (i) Terms of algebraic expression

a2 – b2 – 2ab are: a2, – b2 and – 2ab

(ii) The algebraic expression is 9a2b –4ab2 + abc + 4c

(iii) (a) Coefficient of x in mx is m.

(b) Coefficient of x in – 23 xp is –

23 p.

(iv) – z = (–1)z

The coefficient of z in – z is –1.

(v) The literal factor of 89

xyz

− is

xyz .

2. Numerical coefficient of 3x4 is 3Numerical coefficient of – y3 is – 1Numerical coefficient of z3 is 1Numerical coefficient of 2xyz is 2

Numerical coefficient of – 9 is – 9.

3. (i) 2

2

2

3 8 – 46 – 7

9 7 3

m mm m

m m

++ +

+ +

(ii) 2

2

2 2

2– 3 –

– –

a abab d

a ab d

+

4. (i) 3mn2 + 4m2n2 + (– 5mn2) + (– mn2)

= 3mn2 + 4m2n2 – 5mn2 – mn2

= (3mn2 – 5mn2 – mn2) + 4m2n2

= (3mn2 – 6mn2) + 4m2n2

= 4m2n2 – 3mn2

= mn2 (4m – 3).

(ii) (x – 9) + (3x – 8) + (x – 1)

= x – 9 + 3x – 8 + x – 1

= (x + 3x + x) + (– 9 – 8 – 1)= 5x – 18.

5. Perimeter of the given figure

= (4a – 3b) + (5a – 4b)

+ (4a – 3b) + (5a – 4b)

= (4a + 5a + 4a + 5a)

+ (– 3b – 4b – 3b – 4b)

= 18a – 14b = 2(9a – 7b) units.

6. 2 2

2 2

2

2

2 2

4 – 6– 2

6– 4

3 4 – 8 6

x y xyx y xy

yx xy

x y xy

++

+ ++

+ +

Subtract – 2x2 + y2 – xy + x from 3x2 +4y2 – 8xy + 6.

2 2

2 2

2 2

3 4 – 8 6– 2 –

– – .5 3 – 7 – 6

x y xyx y xy x

x y xy x

+ ++ +

+ ++ +

7. We first add 8b2 – 3c2 and 2b2 + bc– 2c2 as

… (i)

Then, we add 2b2 – 2bc – c2 and c2 +2bc – b2 as

… (ii)

2 2

2 2

2 2

8 – 32 – 210 – 5

b cb bc cb bc c++

2 2

2 2

2

2 – 2 –– 2

b bc cb bc cb

+ +


Now, we subtract the sum (i) from thesum (ii) as

8. (i) 4x – (– 8y – 3x) = 4x + 8y + 3x

= (4x + 3x) + 8y

= 7x + 8y.

(ii) – (49 – a2) – 9a2 = – 49 + a2 – 9a2

= (a2 – 9a2) + (– 49)

= – 8a2 – 49.

WORKSHEET –94

1. Let A should be added. Then

A + m2 + mn + n2 = 3m2 + 4mn

∴ A = 3m2 + 4mn – m2 – mn

= 2m2 + 3mn

= m(2m + 3n).

2. Let S should be subtracted. Then,

3x2 + 9y2 + 10 + 12xy – S

= – x2 – y2 + 12 + 8xy

or 3x2 + 9y2 + 10 + 12xy + x2 + y2 – 12 – 8xy = S

or 4x2 + 10y2 + 4xy – 2 = S

i.e., S = 4x2 + 10y2 + 4xy – 2.

3. We first add 4x – y + 12 and – y + 12 as

4 – 12 – 12

4 – 2 24

x yy

x y

+++

Then we subtract 6x – y – 20 from4x – 2y + 24 as

4 – 2 246 – – 20

– .–– 2 44

x yx y

x y

+

+ ++

4. We first add 8 + 2x and 6 – 6x + 3x2 as

2

2

8 26 – 6 3

14 – 4 3

xx xx x

+++ …(i)

Then, we add 3x2 – 6x and – 2x2 + 2x – 5as

2

2

2

3 – 6– 2 2 – 5

– 4 – 5

x xx x

x x+

…(ii)

Now we subtract the sum (2) from thesum (i) as

2

2

2

3 – 4 14 – 4 – 5

–

2 19

x xx x

x

+

+ ++

Thus, the result is 2x2 + 19.

5. 6x – 3y2 – 8y + y – 3x

= (6x – 3x) – 3y2 + (– 8y + y)

= 3x – 3y2 – 7y.

6. 2 2

2 2

2 2

30 40 – 12– 15 – 30 72

–.45 70 – 84

p q pqp q pq

p q pq

++

+ ++

7. (i) 2x + x = 3x.

(ii) 7y – 3y = 4y.

(iii) 8m + 8m = 16m.

(iv) 12y – 12y = 0.

8. (i) 3a + 5 –13 – 4a = (3a – 4a) + (5 – 13)

= – a + (– 8)

= – a – 8.

2

2 2

2 2

10 – 5

– –

– 9 – 5

bb bc c

b bc c

+++


(ii) 5x – 3y2 – 8y + y – 4x

= (5x – 4x) + (– 3y2) + (– 8y + y)

= x – 3y2 – 7y.

9. (i) 10x – 4x2 – 2x2

Here, – 4x2 and – 2x2 are the liketerms.

∴ 10x – 4x2 – 2x2

= 10x + (– 4x2 – 2x2)

= 10x + (– 6x2)

= 10x – 6x2.

(ii) 17ab – 7ba + 2bc

Here, 17ab and – 7ba are the liketerms.

∴ 17ab – 7ba + 2bc

= (17ab – 7ba) + 2bc

= (17ab – 7ab) + 2bc

(∵ ab = ba)

= 10ab + 2bc.

10. (i) a – (a – b) – b – (b – a)

= a – a + b – b – b + a

= (a – a + a) + (b – b – b)

[∵ a, – a, a are like terms aswell as b, – b, – b are like terms.]

= a – b.

(ii) (4a2 + 5a – 4) – (8a – a2 – 5)

= 4a2 + 5a – 4 – 8a + a2 + 5

= (4a2 + a2) + (5a – 8a) + (– 4 + 5)

[∵ 4a2 and a2 are like terms; 5aand – 8a are like terms; – 4 and 5 are like terms]

= 5a2 – 3a + 1.

WORKSHEET –95

1. (i) 2

2

2

2

62 – 93 10

.11 1

x yx yx yx y

++

(ii) 2 2

2 2

2 2

2 2

– – 3– – 3

– – 3.– 3 – 3

y zy zy z

y z+

2. (i) 8(b – a) = 8b – 8a

6(b – a) = 6b – 6a

Subtract 8b – 8a from 6b – 6a as

6 – 68 – 8

–– 2 2

b ab a

b a++

– 2b + 2a = 2(a – b).

(ii) Subtract 24ab – 10b – 15a from40ab + 16b + 18a as

40 16 1824 – 10 – 15

–.16 26 33

ab b aab b a

ab b a

+ +

+ +++

3. (i) Substituting x = 2 in 2x – 3, we get

2x – 3 = 2 (2) – 3 = 4 – 3 = 1.

(ii) Substituting x = 2 in 4x2 – x – 6, weget

4x2 – x – 6 = 4(2)2 – (2) – 6

= 4 × 4 – 2 – 6

= 16 – 8 = 8.

4. (i) Substituting a = 2 and b = – 1 ina2 – b2 – 4, we get


a2 – b2 – 4 = (2)2 – (–1)2 – 4

= 4 – 1 – 4 = – 1.

(ii) Substituting a = 2 and b = –1 in a – b2 – a2b2, we get

a – b2 – a2b2 = 2 – (– 1)2 – (2)2 (–1)2

= 2 – (1) – (4) (1)

= 2 – 1 – 4 = – 3.

5. (i) Substituting p = 2, a = – 1, b = – 2and q = 0

in 5p – 5q – 5 + a – b, we get

5p – 5q – 5 + a – b

= 5(2) – 5(0) – 5 + (– 1) – (– 2)

= 10 – 0 – 5 – 1 + 2 = 6.

(ii) in pq + ab + pa, we get

pq + ab + pa

= (2) (0) + (– 1) (– 2) + (2) (– 1)

= 0 + 2 – 2 = 0.

6. 3p2 + p – a = 8

or 3(1)2 + (1) – a = 8 (∵ p = 1)

or 3 + 1– a = 8

or 4 – a = 8

or – a = 8 – 4


or – a = 4

∴ a = – 4.

7.∵ x = 0 and y = – 2

∴ 4x2y + 2xy2 – 2xy + 8

= 4(0)2 (– 2) + 2(0) (– 2)2

– 2(0) (– 2) + 8

= 0 – 0 – 0 + 8 = 8.8. Perimeter of triangle

= Sum of the measures of sides.= 3x + 1 + 4x + 2 + 5x= (3x + 4x + 5x) + (1 + 2)

= 12x + 3.

9. Total money spent by Reeta

= Money spent on toys + Moneyspent on books

= 4x + 3y + 7x – 3y

= (4x + 7x) + (3y – 3y)

= 11x + 0 = ` 11x.

10. The remaining length of the wire

= (7x – 3) – (2x – 1)

= 7x – 3 – 2x + 1

= (7x – 2x) + (– 3 + 1)

= (5x – 2) m.

WORKSHEET –96

1. (i) Product of a and b = a × b = ab

Subtracting 7 from ab, we get ab – 7.

So, the required algebraic expressionis ab – 7.

(ii) Difference of x and y = x – y

One-third of x – y = 13 (x – y)

So, the required expression is

13 (x – y).

2. (i) Terms of the expression 4x – 3y are4x and – 3y.

(ii) Terms of the expression 8 – x + yare 8, – x and y.

(iii) Terms of the expression y2x – y arey2x and – y.

(iv) Terms of the expression 2z – 5xz are2z and – 5xz.

3. Groups of the like terms are givenbelow:

(a) 13x, – 25x, x, – 12x

(b) – 25y, 12y, y

(c) 13, – 25, 1


4. (i) 2 2 2

2 2 2

2 2 2

4 – 7 7 – 2– – 8

.3 – 6 – – 2

x x y y xyx x y y

yx x y xy

++

(ii) 4 – 12 – 93 – 3 – 4

.7 – 15 – 13

x zx z

x z

5. a(b – 6) = ab – 6a

b(6 – a) = 6b – ab

Now subtract ab – 6a from 6b – ab.6 – – 6 – +

.6 – 2 6

b abab a

b ab a

+

+6. Let A should be added. Then

A + 2x2 + y2 – xy = 4x2 – 3y2

∴ A = 4x2 – 3y2 – 2x2 – y2 + xy= (4x2 – 2x2) + (– 3y2 – y2) + xy= 2x2 – 4y2 + xy.

7. We first add 2x – y + 12 and – x – 20 as

2 – 12– – 20

– – 8

x yxx y

+

Now subtract 4x – y + 20 fromx – y – 8 as

– – 8 4 – 20

– –– 3 – 28

x yx y

x

+ ++

Thus, the result is – 3x – 28

8. Perimeter of a triangle is the sum ofmeasures of its sides.

23 32 – 66 –

m nm nm nm n

++

Therefore, the perimeter of the triangleis (6m – n) units.

9. As 2x + 4y has two terms, it is abinomial.

10. 7x – 2x + 3y – x + 10y – 4x + 2y

= (7x – 2x – x – 4x) + (3y + 10y + 2y)

[Grouping like terms]

= (7x – 7x) + (15y)

= 0 + 15y = 15y.

11. Since a and b are the algebraic factorsof each of the given terms. Therefore,the given terms are like.

WORKSHEET –97

1. Product of x and y = x × y = xy.

Four times of xy = 4xy.

Adding 7 to 4xy, we get 4xy + 7.

Hence, the required algebraic expre-ssion is 4xy + 7.

2. Terms in ab + 2b2 – 3a2 are ab, 2b2 and– 3a2.

Factors of ab are 1, a and b.

Factors of 2b2 are 2, b and b.

Factors of – 3a2 are – 3, a and a.

3. Numerical coefficient of the term 0.1xis 0.1.

Numerical coefficient of the term 0.01y2

is 0.01.

4. 3x2y – 12xy2 + 8y2 – 7xy

= 3x2y + 8y2 – 12xy2 – 7xy

= 3x2y + 4(2 – 3x)y2 – 7xy

Now the coefficient of y2 in theexpression is same as the coefficient ofy2 in 4(2 – 3x)y2, which is 4(2 – 3x).

5. Groups of like terms are given below:

(a) 13xy, – 7xy, 12xy

(b) 7x, – 200x, – 5x, – 3x


(c) 8y, – 7y, 4y

(d) – x2y2, 12x2y2

6. p – (p – q) – p – (q – p) + q – (p – 2q)

= p – p + q – p – q + p + q – p + 2q

= (p – p – p + p – p) + (q – q + q + 2q)

= (2p – 3p) + (4q – q)

= – p + 3q.

7. Add 4ab, a + b – 3ab and 7ab – b as

4– 3

– 7 + 8

aba b ab

b aba ab

++

Thus, the result is a + 8ab.

8. Subtract 25ab – 12b – 6a from30ab – 14b + 2a as

30 – 14 225 – 12 – 6

– .5 2 8–

ab b aab b a

ab b a

+

+ ++

9. Substituting n = – 3 in n2 – 5n3 + 2n + 3,we get

n2 – 5n3 + 2n + 3

= (– 3)2 – 5(– 3)3 + 2(– 3) + 3

= 9 – 5(– 27) – 6 + 3

= 9 + 135 – 6 + 3

= (9 + 135 + 3) – 6

= 147 – 6 = 141.

10. Substituting a = 3 and b = – 1 in a3 – b3,we get

a3 – b3 = (3)3 – (– 1)3

= 27 – (– 1) = 27 + 1 = 28.

11. We know that lengths of all sides ofan equilateral triangle are equal.

∴ Sum of the sides of the given triangle

= (2x + 3y – 8) + (2x + 3y – 8)

+ (2x + 3y – 8)

= (2x + 2x + 2x) + (3y + 3y + 3y)

+ (– 8 – 8 – 8)

= 6x + 9y – 24.

12. Perimeter of a square = 4 × Side

= 4 × (8x + 4y)

= 32x + 16y.

WORKSHEET –98

1. Substituting p = 3, a = – 1, b = – 2 andq = 0.

(i) In 30 – 3p – 3b + p2, we get

30 – 3p – 3b + p2

= 30 – 3(3) – 3(– 2) + (3)2

= 30 – 9 + 6 + 9

= 30 + 6 – 9 + 9 = 36.

(ii) In 3pq + 4ab + pa, we get

3pq + 4ab + pa

= 3(3)(0) + 4(– 1)(– 2) + 3 (– 1)

= 0 + 4 × 2 – 3

= 8 – 3 = 5.

2. Substituting x = 2 in 8

3x

– 5, we get

83x

– 5 = 8 2

3×

– 5 = 163 – 5

= 16 – 15

3 = 13 .

3. Amount left with Reeshu

= ` (18x2 + 3x – 3) – ` (6x2 – 2x – 1)

= ` (18x2 + 3x – 3 – 6x2 + 2x + 1)

= ` (18x2 – 6x2 + 3x + 2x – 3 + 1)

= ` (12x2 + 5x – 2).


4. (i) 40ab + 16b + 18a – (24ab – 10b – 15a)

= 40ab + 16b + 18a

– 24ab + 10b + 15a

= (40ab – 24ab) + (16b + 10b)

+ (18a + 15a)

= 16ab + 26b + 33a.

(ii) 30p2 + 40q2 – 12pq

– (63pq – 30q2 – 15p2)

= 30p2 + 40q2 – 12pq

– 63pq + 30q2 + 15p2

= (30p2 + 15p2) + (40q2 + 30q2)

+ (–12pq – 63pq)

= 45p2 + 70q2 – 75pq.

5. 2y2 – 5 – 2y + y2 – 3y + 6 – y2 – 4y – y2

+2

= (2y2 + y2 – y2 – y2)

+ (– 2y – 3y – 4y) + (– 5 + 6 + 2)

= (3y2 – 2y2) + (– 9y) + (– 5 + 8)

= y2 – 9y + 3.

6. (i) Add c2 + 2cd and –3cd – d2 as

2

2

2 2

2 – 3 –

– –

c cdcd d

c cd d

+

(ii) Add x2 – y2, 2x2 – 3xy + 4y2 and 3y2

– 5xy – x2 as

2 2

2 2

2 2

2 2

– 2 4 – 3

– 3 – 5

2 6 – 8

x yx y xyx y xy

x y xy

+++

7. Polynomial Degree

m2n3 + mn2 + 4 5

ab + bc + ca 2

8. cab2, b2ac and acb2 have same factorswhich are c, a, b and b.

a2bc, c2ab and abc have different factors

Therefore, cab2, b2ac and acb2 are liketerms.

9. Add 9p + 3q, 4p – q and 2p – 2q as9 34 –2 – 2

15

p qp qp q

p

+

Therefore, the total cost is ` 15p.

10. (i) Sum of p and q = p + q.

One-fourth of (p + q) = 14 (p + q)

Therefore, algebraic expression is14 (p + q).

(ii) Product of s and t = s × t = st.

Subtracting 20 from st, we get

st – 20.

Therefore, algebraic expression isst – 20.

(iii) Sum of x and y = x + y.

Product of x and y = x × y = xy.

Subtracting (x + y) from xy, we getxy – (x + y) = xy – x – y

= xy – (x + y).

Therefore, algebraic expression is = xy – (x + y).

❏❏

163I MOPXE EN TN S A DN P EWO R S

WORKSHEET –99

1. (D) 100000 = 10 × 10 × 10 × 10 × 10

= 105.

2. (C)106 is read as ‘10 raised to the power6’.

3. (A) In mn, m is the base and n is theexponent.

4. (A) 128 = 2 × 2 × 2 × 2

× 2 × 2 × 2

= 27.

5. (B)(– 3)3 × (– 5)5

= (– 3 ) × (– 3 ) × (– 3 ) × (– 5 )

× (– 5 ) × (– 5 )× (– 5 )× (– 5 )

= 84375.

6. (C) (a × a) × (b × b × b) × (c × c × c × c)

= a2 × b3 × c4

7. (A) 3600 = 2 × 2 × 2 × 2

× 3 × 3 × 5 × 5

= 24 × 32 × 52

8. (B) Let us take option (B).

28 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2

14Chapter

EXPONENTS AND POWERS

= 256

82 = 8 × 8 = 64

∵ 256 > 64 ∴ 28 > 82.

9. (D) 32 × 35 = 32+5 = 37.

10. (C) 1012 ÷ 103 = 12

31010

= 1012 – 3 = 109.

11. (D) (x5)y = x5 × y = x5y.

12. (C) –5 3

22 5 7

5 7× ×

×= – 2 ×

5

255

× 37

7= – 2 × 55 – 2 × 73 – 1

= – 2 × 53 × 72.

13. (B)2 7

3 436 6

12t

t× ×

×=

(7 – 4)36 6 612 12 12

t× × ×× ×

= 33

4t

.

14. (B) (–1)even number = 1

(–1) odd number = –1.

15. (A) 1000000 = 1 0 × 1 0 × 1 0 × 1 0 × 1 0 × 1 0

= 106.

∴ Standard form of 1000000

= 1.0 × 106.

16. (C) 602800000000000 m3

= 6028 × (10 × 10 × 10 × 10 × 10

× 10 × 10 × 10 × 10 × 10 × 10) m3

= 60281000

× 1000 × 1011 m3

= 6.028 × (10 × 10 × 10) × 1011 m3

= 6.028 × 103 × 1011 m3

= 6.028 × 1014 m3.

2 1282 642 322 162 82 42 2

1

2 36002 18002 9002 4503 2253 755 255 5

1


17. (A) 815 ÷ 810 = 15

1088

= 815 – 10 = 85.

18. (B) 5x × 56 = 5x + 6 = 56 + x.

19. (D) Let us take option (D).

RHS = (xy)3 = (x3y3) = x3 × y3 = LHS.

20. (B)675 = 3 × 3 × 3 × 5 × 5

= 33 × 52

= 33 × (– 1)2 52

= 33 × (– 1 × 5)2

= 33 × (– 5)2.

21. (A) ∵ (2 – 1 – 3 – 1)– 1

= –1 –11

2 – 3=

11 1

–2 3

=1

3 – 26

= 6.

And (2 – 1 + 3 – 1) – 1

= –1 –11

2 3+ = 1

1 12 3+

= 1

3 26+

= 65

∴(2–1 – 3–1)–1 + (2–1 + 3–1)

= 6 + 65 =

30 65+

= 365 .

22. (B) Let the required number be x. Then

x × 2– 1 = 1 or x × 12 = 1

∴ x = 1 × 2 = 2.

23. (D) 4 × 2x + 2 = 2 or 2x + 2 = 24 =

12

or 2x + 2 = 2 – 1

Comparing exponents of 2 on both thesides, we get

x + 2 = – 1

∴ x = – 1– 2 = – 3.

WORKSHEET – 100

1. (i) 25 = 2 × 2 × 2 × 2 × 2 = 32

(ii) 93 = 9 × 9 × 9 = 729

(iii) 84 = 8 × 8 × 8 × 8 = 4096.

2. (i) In 63, 6 is the base and 3 is theexponent. Expanded form of 63.

= 6 × 6 × 6 = 216

= 2 × 100 + 1 × 10 + 6 × 1

= 2 × 102 + 1 × 101 + 6 × 100

(ii) In 82, 8 is the base and 2 is theExponent.Expanded form of 82

= 8 × 8 = 64= 6 × 101 + 4 × 100.

3. (i) m × m = m2.(ii) 4 × 4 × x × x = (4 × 4) × (x × x)

= 42 × x2.(iii) 5 × 5 × 3 × 3 × 3 × 3

= (5 × 5) × (3 × 3 × 3 × 3)= 52 × 34.

4. (i) 22 × 32 = 4 × 9 = 36.

(ii) ( )334

= 34

×34

×34

= 3 3 34 4 4× ×× × =

2764

.

(iii) 32 × 104 = 3 × 3 × 10 × 10 × 10 × 10

= 9 × 10000 = 90000.

5. (i) (–1)3 = (–1) × (–1) × (–1)

= 1 × (–1) = – 1.

(ii) (– 4)2 × (– 5)2 = 16 × 25 = 400.

(iii) (– 2)3 × (– 3)2 = (– 2)3 × (3)2

= – 8 × 9 = – 72.

6. (i)18

= 1

2 2 2× × = 31

2= 1 × 2– 3 = 2– 3

= ( )312

3 6753 2253 755 255 5

1


(ii) 1

64−

= 6– 12

= – 1 × 2 – 6

= – 2 – 6.

(iii)4981

= 7 7

3 3 3 3×

× × ×

= 2

473

= 72 × 3 – 4.

7. (i) x × x3 × x10 = x1 × x3 × x10

= x1 + 3 + 10 = x14.

(ii) (72)3 = 72 × 3 = 76.

(iii) (2016 ÷ 2013 ) × 203 = 2016 – 13 × 203

= 203 × 203

= 203 + 3 = 206.

8. (i)6

3xx

× x5 = x6 – 3 × x5

= x3 × x5 = x3 + 5 = x8.

(ii) (20 + 30) × 40= (1 + 1) × 1

(∵ a0 = 1 for a ! 0)

= 2 × 1 = 21.

9. (i) a3 × b3 = (a × b)3 = (ab)3.

(ii) 34 × 54 = (3 × 5)4 = 154.

10 (i) 20068 = 2 × 10000 + 0 × 1000 + 0

× 100 + 6 × 10 + 8 × 1

= 2 × 104 + 0 + 0 + 6 × 101

+ 8 × 100

= 2 × 104 + 6 × 101 + 8 × 100.

(ii) 176428 = 1 × 100000 + 7 × 10000

+ 6 × 1000 + 4 × 100 + 2

× 10 + 8 × 1

= 1 × 105 + 7 × 104 + 6 × 103

+ 4 × 102 + 2 × 101

+ 8 × 100.

WORKSHEET – 101

1. (i) 3 × 103 + 101 + 4 × 100

= 3 × 1000 + 10 + 4 × 1 (∵ 100 = 1)

= 3000 + 10 + 4 = 3014.

(ii) 4 × 105 + 3 × 102 + 2 × 10 + 100

= 4 × 100000 + 3 × 100 + 2 × 10 + 1

= 400000 + 300 + 20 + 1 = 400321.

2. (i) 4985.5 = 4.9855 × 1000

= 4.9855 × 103.

(ii) 4450000 = 4.450000 × 1000000

= 4.45 × 106.

3. (i) Population = 640800

= 6.40800 × 105

= 6.408 × 105.

(ii) The literate population of India

= 37000000 = 3.7000000 × 107

= 3.7 × 107.

4. (i) 59853 × 103

= 59853 × 1000 = 59853000

= 5 × 10000000 + 9 × 1000000

+ 8 × 100000 + 5 × 10000 + 3

× 1000

= 5 × 107 + 9 × 106 + 8 × 105

+ 5 × 104 + 3 × 103.

(ii) 7644 × 10 – 5

= (7 × 1000 + 6 × 100 + 4 × 10

+ 4 × 1) × 10 – 5

= (7 × 103 + 6 × 102 + 4 × 101 + 4 × 100) × 10 – 5

2 642 322 162 82 42 2

1

3 813 273 93 3

1


= 7 × 103 – 5 + 6 × 102 – 5 + 4 × 101 – 5

+ 4 × 100 – 5

= 7 × 10 – 2 + 6 × 10 – 3 + 4 × 10 – 4

+ 4 × 10–5.

5. (i) x × x × x × x × a × a

= (x1 × x1 × x1 × x1) × (a1 × a1)

= x1+1+1+1 × a1+1 = x4 × a2 = x4a2.

(ii) 4 × 4 × 4 × 6 × 6 × 6

= (41 × 41 × 41) × (61 × 61 × 61)

= 41+1+1 × 61+1+1 = 43 × 63

= (4 × 6)3 = 243.

6. (i) 23 × 5 = (2 × 2 × 2) × 5 = 8 × 5 = 40.

(ii) 32 × 102 = (3 × 3) × (10 × 10)

= 9 × 100 = 900.

7. (i) ( )425

= 4

425

= 2 2 2 25 5 5 5× × ×× × ×

= 16625

= 1 10 6 1

6 100 2 10 5 1× + ×

× + × + ×

= 1 0

2 1 01 10 6 10

6 10 2 10 5 10× + ×

× + × + ×.

(ii) ( )345

= 3

345

= 4 4 45 5 5× ×× ×

= 64

125

= 6 10 4 11 100 2 10 5 1

× + ×× + × + ×

= 1 0

2 1 06 10 4 10

1 10 2 10 5 10× + ×

× + × + ×.

8. 72 × a2 × 2a5 = 49 × a2 × 2 × a5

= (49 × 2) × (a2 × a5)

= 98 × a2 + 5 = 98a7.

9. 8000000 = 8 × 10 × 10 × 10 × 10 × 10

× 10

= 8 × 106.

10. 200072 = 2 × 100000 + 0 × 10000 + 0 × 1000 + 0 × 100

+ 7 × 10 + 2 × 1= 2 × 105 + 0 + 0 + 0 + 7 × 101

+ 2 × 100

= 2 × 105 + 7 × 101 + 2 × 100.

11. (i) 121169

= 11 1113 13

××

= 2

21113

= ( )21113 .

(ii)– 136

= –1

2 2 3 3× × ×

= – 2 21

2 3× = – 2

1(2 3)×

= – 21

6 = – ( )21

6.

(iii)16625

= 2 2 2 25 5 5 5× × ×× × ×

= 4

425

= ( )425

.

WORKSHEET – 102

1. 256 = 2 × 2 × 2 × 2

× 2 × 2 × 2 × 2

= 28.

2. 25 = 2 × 2 × 2 × 2 × 2 = 32

42 = 4 × 4 = 16

∵ 32 > 16

∴ 25 > 42

Thus, 25 is greater.

5 6255 1255 255 5

1

2 2562 1282 642 322 162 82 42 2

1


3. x3y2 = x × x × x × y × y …(i)

y2x3 = y × y × x × x × x

or y2x3 = x × x × x × y × y …(ii)

From equations (i) and (ii), it is clearthat x3y2 and y2x3 are same.

4. 72 = 2 × 2 × 2 × 3 × 3

= 23 × 32.

5. (– 3)3 = (– 3) × (– 3) × (– 3)

= – (3 × 3 × 3) = – 27

6. 3 × 3 × 3 × x × x × x = 33 × x3.

7. (i) 24 × 34

= (2 × 2 × 2 × 2) × (3 × 3 × 3 × 3)

= 16 × 81 = 1296.

(ii) (– 3) × (– 2)3 = (– 3) × (– 2) × (– 2)

× (– 2)

= 6 × 4 = 24.

8. (i) 43 × 23 is in exponential form.

Let us convert it into expandedform.

43 × 23 = 4 × 4 × 4 × 2 × 2 × 2

= 64 × 8 = 512

= 5 × 100 + 1 × 10 + 2 × 1

= 5 × 102 + 1 × 101 + 2 × 100.

(ii) p5 ÷ q5 is in exponential form.

Let us convert it into expandedform.

p5 ÷ q5 = 5

5

pq

= 5p

q

= pq

× pq ×

pq

× pq ×

pq

.

9. ( )2413

= 4

13 ×

413

= 2 213×

× 2 213×

= 2 2 2 2

13 13× × ×

× .

10. 4650000 = 4.650000 × 106

= 4.65 × 106.

11. In b5, b is the base and 5 is theexponent.

12.649

= 2 2 2 2 2 2

3 3× × × × ×

×

= 6

223

.

13. (i) 32 = 3 × 3 = 9; 23 = 2 × 2 × 2 = 8

∵ 9 > 8 ∴ 32 > 23

So, 32 is greater.

(ii) 25 = 2 × 2 × 2 × 2 × 2 = 32;

52 = 5 × 5 = 25

∵ 32 > 25 ∴ 25 > 52

So, 25 is greater.

WORKSHEET – 103

1. (i) (– 2)2 × 34

= (– 2) × (– 2) × 3 × 3 × 3 × 3

= 4 × 81 = 324.

(ii) (– 1)2 × (– 2)3 × (– 5)

= (– 1) × (– 1) × (– 2) × (– 2)

× (– 2) × (– 5)

= 1 × 4 × 10 = 40.

(iii) ( )223

× ( )212

2 642 322 162 82 42 2

1

2 722 362 183 93 3

1


= 23 ×

23 ×

12 ×

12 =

2 22 2×× ×

13 3×

= 1 × 19 =

19 .

2. 500 = 2 × 2 × 5 × 5 × 5

= 22 × 53.

3. (i) p × p × p × p × p

= p1 + 1 + 1 + 1 + 1 = p5.

(ii) (– 3) × (– 3) × (– 3) × (– 3)

= (– 3)4 = (– 1 × 3)4 = (– 1)4 × 34

= 1 × 34 [∵ (– 1)even number = 1]

= 34.

4. (i) 83 = 8 × 8 × 8 = 64 × 8 = 512.

(ii) ( )3– 37

= – 37

× – 37 ×

– 37

= (– 3) (– 3)

7 7××

× – 37

= 949 ×

– 37 =

– 27343 .

5. a3b2 = a × a × a × b × b … (i)

a2b3 = a × a × b × b × b … (ii)

b2a3 = b × b × a × a × a

= a × a × a × b × b …(iii)

b3a2 = b × b × b × a × a

= a × a × b × b × b …(iv)

From equations (i), (ii), (iii) and (iv), itis clear that they all are not same.

6. (i) 625 = 5 × 5 × 5 × 5

= 54.

(ii) 3125 = 5 × 5 × 5 × 5 × 5

= 55.

7. (i) Substituting x = – 25 in (5x)3, we

get

(5x)3 = ( )3– 25

5× = (– 2)3

= (– 2) × (– 2) × (– 2)

= 4 × (– 2)= – 8.

(ii) Substituting a = 2 and b = – 1, in(– ab), we get

(– ab)= – ( 2) × (– 1) = 2.

8. (i)7

12(– 2)(– 2)

= (– 2)7 – 12 = (– 2)– 5 = 51

(– 2)

= 1

(– 2) (– 2) (– 2) (– 2) (– 2)× × × ×

= 1

4 4 (– 2)× × = 1

16(– 2)

= 1

– 32 =

– 132

.

(ii) (– 4)6 ÷ (– 4)8 = (– 4)6 – 8 = (– 4)– 2

= 21

(– 4)=

1(– 4) (– 4)×

= 1

16 .

9. (i) x3 = 125343

= 5 5 57 7 7× ×× ×

= 3

357

or x3 = ( )357

∴ x = 57

5 6255 1255 255 5

1

5 31255 6255 1255 255 5

12 5002 2505 1255 255 5

1


(ii) (x2)3 = 1

64

orx2 × 3 = 1

2 2 2 2 2 2× × × × ×

or x6 = 612

or x6 = ( )612

∴ x = 12

.

10. Let the required rational number be x.Then,

x × ( )– 123

= – 32

or x × 32

= – 32

∴ x = – 32 ×

23 = –1.

WORKSHEET – 104

1. (i)5

7(– 4)(– 4)

= (– 4)5 – 7 = (– 4)– 2 = 21

(– 4)

= 1

(– 4) (– 4)× =

116

.

(ii)6– x

y = { }6

(–1)xy

× = (–1)6 × 6x

y

= 6x

y

[∵ (–1) even number = 1]

= 6

6xy

.

2. (i) (– 2 × 103)2 = (– 2)2 × (103)2

= (– 2) × (– 2) × 103 × 2

= 4 × 106

= 4 × 1000000

= 4000000.

(ii) 37 × ( )713

= 37 × 7

713

= 17 = 1.

3. a0 = 1 for a ! 0

(i) 30 + 40 + 50 = 1 + 1 + 1= 3.

(ii) (90 – 70) × (9 + 7) = (1 – 1) × 16

= 0 × 16 = 0.

4. (i)8 7

3 45

(25)b

b××

= 8 7

2 3 45

(5 )b

b××

= 8

655

× 7

4bb

= 58 – 6 × b7 – 4

= 52 × b3 = 5 × 5 × b3

= 25b3.(ii) (290 – 230) × 160

= (1 – 1) × 1 (∵ a0 = 1 for a ! 0)= 0 × 1 = 0.

5. (i) 185000 = 185 × 1000= 1.85 × 100 × 1000= 1.85 × 105.

(ii) 400000 = 4.00000 × 100000= 4.0 × 105.

6. 43 = 4 × 4 × 4 = 64 and 52 = 5 × 5 = 25∵ 64 ! 25

Therefore, 43 is not equal to 52.

7. (i) – 216

= – (2 × 2 × 2 × 3 × 3 × 3)

= – (23 × 33)

= – (2 × 3)3

= – 63

= (– 6)3.

(ii)–1243 =

–13 3 3 3 3× × × ×

= 5

5

(–1)3

= ( )5–13

.

2 2162 1082 543 273 93 3

1

2 642 322 162 82 42 2

1

3 2433 813 273 93 3

1


8. Substituting x = –1, a = 2 and y = 1 inx5y2a3, we get

x5y2a3 = (– 1)5 × (1)2 × (2)3

= – 1 × 1 × 8 = – 8.

[∵ (– 1)odd number = – 1]

9. (– 5)6 ÷ (– 5)8 = 6

8

(– 5)(– 5)

= (– 5)6 – 8

= (– 5) – 2 = 21

(– 5)

= 1

(– 5) (– 5)× =

125

.

10. (i)3 2 2

2(3 ) 5

9 5××

= 3 2 2

23 5

(3 3) 5

× ×× ×

= 6 2

2 23 5

(3 ) 5××

= 6 2

43 53 5××

= 6

433

× 25

5

= 36 – 4 × 52 – 1 = 32 × 5

= 9 × 5 = 45.

(ii) (190 – 170) × (19 + 17) = (1 – 1) × 36

[∵ a0 = 1 for a ! 0]

= 0 × 36 = 0.

WORKSHEET – 105

1. (i) (– 6) × (– 6) × (– 6) × (– 6)

= (– 1 × 6)4

= (– 1)4 × 64

= 1 × 64 [∵ (– 1)even number. = 1]

= 64.

(ii) x × x × x × x × x = x1 + 1 + 1 + 1 + 1 = x5.

2. (i) (– 2)4 × (– 2)11 = (– 2)4 + 11 = (– 2)15

(ii) (– 7 )2 × (– 7)11 × (– 7)

= (– 7)2 + 11 + 1 = (– 7)14 = (– 1 × 7)14

= (– 1)14 × 714 = 714.

3. (i) 4x × 42 = 4x + 2.

(ii) 34a × 33a = 34a + 3a = 37a.

4. (i) (42)4= 42 × 4 = 48 = (2 × 2)8

= (22)8 = 22 × 8 = 216 = 65536.

(ii) (52)5= 52 × 5 = 510 = 9765625.

5. (i) (23)5 × (27)2 = 23 × 5 × 27 × 2

= 215 × 214 = 215 + 14

= 229.

(ii) ( ) ( )32 4– 2 2

5 5

×

= ( ) ( )32 42 2

– 15 5

× ×

= { } ( )342

2 2 2(– 1)

5 5

× ×

= ( ) ( )32 42 2

5 5

× = ( )

32 425

+

= ( )362

5

= ( )6 325

×= ( )182

5.

6. ( )32411

× ( )3118

= ( )324 1111 8

×

[∵ a3 × b3 = (a × b)3]

= 33 = 3 × 3 × 3 = 27.

7. Substituting x = – 18 in (8x)3, we get

(8x)3 = ( ) 3–18

8 × = ( )3– 8

8 = (– 1)3

= – 1 [∵ (– 1) odd number. = – 1].

8. (i)6

444

= 46 – 4 = 42 = 4 × 4 = 16.


(ii) ( )4–14

÷ ( )2–14

= ( )( )

4

2

–14

–14

= ( )4 – 2–14

= ( )2–14

= – 14 ×

– 14

= 1

16.

9. (i) (80 – 70) × (8 + 7) = (1 – 1) × 15

(∵ a0 = 1 for a ! 0 )

= 0 × 15 = 0.

(ii) 40 + 30 + 10 = 1 + 1 + 1 = 3.

10. (i)8 7

2 55

(25)b

b××

= 8

2 25

(5 ) ×

7

5bb

= 8

2 25

5 × × b7 – 5 = 8

455

× b2

= 58 – 4 × b2 = 54 × b2

= 5 × 5 × 5 × 5 × b2

= 25 × 25 × b2 = 625 b2.

(ii)( ) ( )( ) ( )

3 3

2 4

3 15 73 15 7

×

×=

( )( )

3

2

3535

× ( )( )

3

4

1717

= ( )3 – 235

× ( )3 – 417

= ( )135

× ( )– 117

= 35 ×

71 =

215 = 4

15 .

11. (i) 154034

= 1 × 100000 + 5 × 10000 + 4

× 1000 + 0 × 100 + 3 × 10 + 4

× 1

= 1 × 105 + 5 × 104 + 4 × 103

+ 0 + 3 × 101 + 4 × 100

= 1 × 105 + 5 × 104 + 4 × 103

+ 3 × 101 + 4 × 100

(ii) 5400500

= 5 × 1000000 + 4 × 100000

+ 0 × 10000 + 0 × 1000 + 5

× 100 + 0 × 10 + 0 × 1

= 5 × 106 + 4 × 105 + 0 + 0 + 5

× 102 + 0 + 0

= 5 × 106 + 4 × 105 + 5 × 102.

❏❏


WORKSHEET – 106

1. (B) A cube can be made by using thenet given in the option (B).

2. (B) A cone has only one vertex.

3. (A) Number of curved edges = a = 2

Number of circular faces = b = 2

Number of curved faces = c = 1

4. (B) The right matching is

(a) → (iii), (b) → (i), (c) → (ii).

5. (D) A cube and a cuboid have equalnumber of edges, i.e., 12.

6. (A) The given figure is of a cone.

7. (C) Cutting horizontally the pipe (seeFig. (i)), the cross section obtained is acircle (see Fig. (ii)).

8. (B) A cuboid has 6 faces.

9. (D) Number of cubes

= 4 × 4 × 2 = 32.

10. (A) A two-dimensional(2-D) sketch of a cubemay be as follows:

15Chapter

VISUALISING SOLID SHAPES

11. (B) A cube has 8 vertices.

12. (C) A cylinder has a curved face andtwo flat faces.

13. (B) A 2-D shape of a cone may be asgiven below:

14. (D) The given net corresponds to acube.

15. (D) A brick is in the form of a cuboidwhich has 8 vertices.

16. (A) A circular pipe is a cylinder.

WORKSHEET – 107

1. (i) Shape → cone

Number of edges → 1

Number of faces → 2

Number of vertices → 1

(ii) Shape → Cuboid

Number of edges → 12

Number of faces → 6

Number of vertices → 8.

2. Net for a cylinder:

173U LA I NS G O DIL S PAH EIV S I S S

3. 4.

5. When a horizontal cut is given to adie, a square cross-section is obtained.

6. Net for a cone.

7.



8. (i) Square (ii) Rectangle (iii) Circle.

9. (i) Cube (ii) Cone.

10. (i) Rectangle (ii) Rectangle.

WORKSHEET – 108

1. Cube

2. (i) Cuboid (ii) Sphere

3. Circle

4.

5.

6. Rectangle.

7. Four.


8.

Length of the new figure

= 3 cm + 3 cm = 6 cm

Breadth of the new figure = 2 cm

Height of the new figure = 2 cm.

9.

Length of the new figure = 8 cm

Breadth of the new figure

= 3 cm + 3 cm = 6 cm

Height of the new figure = 3 cm.

10.

11. Number of cubes = 4.

12. A cylinder has three faces

WORKSHEET – 109

1.

Length of the resulting cuboid

= 3 cm + 3 cm = 6 cm

Breadth of the resulting cuboid = 3 cm

Height of the resulting cuboid = 3 cm.

2. Length = 4 cm + 4 cm + 4 cm = 12 cm

Breadth = 4 cm

Height = 4 cm


3. (i) Cone (ii) Cone (iii) Cone.

4.

5. (i) Circle (ii) Triangle

6.

7.

8. (i) Cylinder (ii) Cone

9. (i) Cube:

(ii) Sphere:

(iii) Cylinder:

10. Square.

Oblique sketch

Oblique sketch


WORKSHEET – 110

1. (i) Squares (ii) Triangles.

2. On folding up the given net, you donot get a cube.

3.

4.

5.

6. (i) (ii)

7. (i) Cone (ii) Pyramid

8. (i) Tetrahedron (ii) Tetrahedron

(iii) Cylinder (iv) Triangular prism

(v) Sphere (vi) Cuboid.

9. In a cube,

number of faces = 6

number of edges = 12

WORKSHEET – 111

1.

2. (i) Cube(ii) Triangular prism

3. (i) Triangles and parallelograms(ii) Circles and rectangle

4. Yes, the given net could form apyramid.

5.

6. (i) Circle (ii) Rectangle(iii) Triangle

7. (i) A vertical cut gives a rectangularcross-section. A horizontal cut givesa circular cross-section.

(ii) Both the vertical and horizontal cutsgive a squared cross-section.

(iii) Both the vertical and horizontal cutsgive a rectangular cross-section.

8. (i) Top view(ii) Front view

(iii) Side view. ❏❏

179C EP R C TA I P A E RP S

PRACTICE PAPERS

Practice PapePractice PapePractice PapePractice PapePractice Paperrrrr ––––– 11111

SECTION-A1. (C) 6 – (– 4) = 6 + 4 = 10.2. (C) A mixed fraction is a combination

of a whole number and a properfraction.

3. (B) 44.89 + x = 80.25x = 80.25 – 44.89

∴ = 35.36.

4. (A) Mean = 46 45 60 54 70 496

+ + + + +

= 3246

= 54.

5. (A) One-fourth of m =4m

∵4m

is 3 more than 7.

∴ 4m

– 3 = 7 or 4m

– 7 = 3.

6. (B) We get an endless line segmentwhich is called a line.

7. (B) Two line segments are congruentif they have same length.

8. (A) 3 km 3000 m300 m 300 m

= = 10 : 1.

9. (C) 5

– 3 =

5– 3

× 44

= 20

–12

10. (A) r = 30 cmArea = pr2

= 227

× 30 × 30 = 19800

7= 2828.57 ¡ 2828 cm2.

SECTION-B

11. All integers between – 3 and 3 are:– 2, – 1, 0, 1, 2.

12. (– 2 – 5) × (– 6) = (– 7) × (– 6) = 42(– 2) – 5 × (– 6) = – 2 + (– 5 × (– 6))

= – 2 + 30 = 28∵ 42 > 28∴ (– 2 – 5) × (– 6) is greater than (– 2)– 5 × (– 6).

13.29

÷ 12

= 29

× 21

= 49

Reciprocal of 29

÷ 12

= Reciprocal of49

=94

.

14. ∵ 1 = 13

+ 13

+ 13

∴ 3 = 1 1 13 3 3

+ + × 3

= (3 one-thirds) × 3= 9 one-thirds

Thus, there are 9 one-thirds in 3.15. ∵ xyz = yzx = zxy

∴ 6xyz – 10yzx + 12zxy

= 6xyz – 10xyz + 12xyz.

= 18xyz – 10xyz

= 8xyz.16. 6x + 14 = 16

Subtracting 14 from both sides, weget 6x = 2.Dividing both sides by 6, we get

180

x = 26

= 13

.

Thus, x = 13

.

17. 36 : 81 = 3681

= 9 49 9

××

= 49

= 4 : 9.

18. CP = SP + Total loss = ` 18 + ` 2 = ` 20.

Loss per cent = Total loss

CP × 100

= 220

× 100

= 10010

= 10.

Thus, the loss per cent is 10%.

SECTION -C

19. Total CP for Renu = 7500 + 500= ` 8000

Loss for Renu = 12% of CP

= 12100

× 8000

= ` 960SP for Renu = CP – Loss

= 8000 – 960= ` 7040

Now, CP for Deepa = SP for Renu= ` 7040.

Thus, the cost price of the T.V. forDeepa is ` 7040.

20. P = ` 750.50, T = 3 years, R = 12%

I = PRT100

= 750.50 12 3

100× ×

= 75050100

× 36

100 =

270180010000

= ` 270.18

A = P + I = ` 750.50 + ` 270.18= ` 1020.68.

Thus, the simple interest is ` 270.18and the amount is ` 1020.68.

21. Let the given angles be 4A and 6A.∵ Sum of the two supplementary

angles = 180°.∴ 4A + 6A = 180°or 10A = 180°∴ A = 18°∴ 4A = 4 × 18° = 72°and 6A = 6 × 18° = 108°.Thus, the required angles are of 72°and 108°.

22. Let the given angles be 7x and 2x.

In ∆ ABC,90° + 2x + 7x = 180°

(Angle sum property of a triangle)or 90° + 9x = 180°∴ 9x = 180° – 90° = 90°

or x = 909°

= 10°

(Dividing both sides by 9)∴ 7x = 7 × 10° = 70°and 2x = 2 × 10° = 20°.Thus, the angles are of measures 70°and 20° respectively.

23. AB y CD and BC is transversal∴ ∠B = ∠BCD

= 90°In ∆ABC,

∠A + ∠B + x = 180° (Angle sum property of a triangle)or 55° + 90° + x = 180°

AM T H E M A T C SI VII–


or 145° + x = 180°∴ x = 180° – 145° = 35°Thus, the value of x is 35°.

24. Percentage of marks

= Marks obtained

Maximum marks × 100

Ravi:Obtained marks = 850

Maximum marks = 900

∴ Percentage of marks =850900

× 100

= 850

9= 94.44%

Rohit:Obtained marks = 540

Maximum marks = 600

∴ Percentage of marks =540600

× 100

= 5406

= 90%

Since Ravi obtained more percentageof marks. Therefore, Ravi’s perform-ance is better.

25. ∵23

4 =

34

× 34

= 3 × 34 × 4

= 9

16,

3– 12

= – 12

× – 12

× – 12

= 12

× 12

× – 12

= – 18

and 23 = 2 × 2 × 2 = 8

∴ 23

4 ×

31–

2 × 23

= 9

16 ×

1–

8 × 8

= – 9

16 × (– 1) = –

916

.

26. 4.346 – 1.16 + 3.402 – 2.3= 4.346 + 3.402 – 1.16 – 2.3= (4.346 + 3.402) – (1.16 + 2.3)= 7.748 – 3.46= 4.288.

27.

When we give a vertical cut to a brick,we get a rectangular cross-section.

28. A die has six faces marked withnumbers 1 to 6, one number on oneface.All possible outcomes are 1, 2, 3, 4, 5and 6.∴ Total number of possible outcomes= 6

(i) Probability

= Number of 3’s

Total number of possible outcomes

= 16

(ii) Probability

= Number of 6’s

Total number of possible outcomes

= 1

.6

SECTION -D

29. In order to construct a bar graph, youhave to go to the following steps:

Step I. Take a graph paper and drawa pair of perpendicular lines OX andOY. Call OX as the horizontal axisand OY as the vertical axis.

Step II. Along OX, mark the namesof the given students and choose the

182

equal width of the bars and uniformgap between them.

Along OY, mark the marks obtainedby the students.Step III. Choose a suitable scale ony-axis to determine heights of thebars. You can choose. 1 big division = 50 marks.

Step IV. Calculation for heights ofvarious bars:

Height of the bar for Romi

= 45050

= 9 big divisions

Height of the bar for Neetu

= 30050

= 6 big divisions

Height of the bar for Ria

= 46050

= 9.2 big divisions

Height of the bar for Lata

= 40050

= 8 big divisions

Height of the bar for Sony

= 34050

= 6.8 big divisions

Step V. Draw the bars with heightsobtained in the step IV and write the



corresponding marks of the studenton the top of each bar.

30.(i) One-fifth of y = 5y

Subtracting 7 from 5y

, we get

5y

– 7

According to the given condition,

5y

– 7 = 7

This is the required equation.Transposing – 7 to RHS, we get

5y

= 7 + 7

or5y

= 14


5y

× 5 = 14 × 5 or y = 70.

(ii) One-fourth of x = 4x

Adding 4 to 4x

, we get 4x + 4.

According to the given condition.

4x

+ 4 = 30

This is the required equation.Transposing 4 to RHS, we get

4x

= 30 – 4 or 4x

= 26.

Multiplying both sides by 4, we get x = 104.

31.(i)– 8– 13

+ 31

– 39 +

– 1126

+ 3

= 8

13 +

31

– 31 1139 26

+

= 8 3 13

13+ ×

– 2 31 11 3

78× + ×

= 8 39

13+

– 62 33

78+

= 4713

– 9578

= 47 6 – 95 1

78× ×

= 282 – 95

78

= 18778

= 23178

.

(ii) – 125

+ 3110

+ 11

– 15 +

– 7– 20

= 31 710 20

+ – 12 115 15

+

= 31 2 7 1

20× + ×

– 12 3 11 1

15× + ×

= 62 720+ –

36 1115+

= 6920

– 4715

= 69 3 – 47 4

60× ×

= 207 – 188

60 =

1960

.

32.(i) Side of the given square = 22 cm.Area of the square = Side × Side

= 22 × 22 = 484 cm2

Radius of the given circle, r = 3 cmArea of the circle = pr2

= 227

× 3 × 3

= 1987

cm2

Area of the shaded portion= Area of the square – Area of the circle

= 484 – 1987

= 484 7 – 198

7×

= 3388 – 198

7 =

31907

.

= 455.71 cm2.

184

(ii) Area of the big rectangle= Length × Breadth= 25 × 20 = 500 m2.

Area of the small rectangle= Length × Breadth= 10 × 5 = 50 m2.

Area of the shaded portion= Area of the big rectangle – Area of the small rectangle.= 500 – 50 = 450 m2.

33.(i) (ii)

Net for a cone. Net for a cylinder

(iii)

(iv)

(v)

34. Radius of circle, r = 21 cmPerimeter of the square

= circumference of the circle= 2pr

= 2 × 227

× 21 = 44 × 3 = 132 cm

Let side of the square = a.Then,

4a = 132

∴ a = 132

4= 33 cm

Area of the square = a2

= 33 × 33= 1089 cm2.

Practice Paper –2

SECTION-A1. (C) Additive inverse of – 200 is 200.

2. (C) 513

÷ 443

= 163

÷ 163

= 1.

3. (D) 8 – x = 3.187∴ x = 8 – 3.187 = 8.000 – 3.187

= 4.813.

4. (B) x – 32

= 5 or x = 32

+ 5 = 132

.

5. (A) A closed figure bounded by threeline segments is called a triangle.

6. (C) Measurements of two congruentangles are equal.

∴ m∠p = m∠q.

7. (A) 150% = 150100

= 50 350 2

××

= 32

.

8. (C) There are infinitely many rationalnumbers between – 2 and – 1. One

of them is – 3

.2

9. (D) A parallelogram has no lines ofsymmetry.

10. (A) The expression ‘a2 + b2’ has 2 termsand so it is a binomial.



SECTION-B

11. Let the number be x.Then according to question,

3x + 11 = 32 (Transposing 11 to RHS)

or 3x = 32 – 11 = 21

∴ x = 213

= 7.

12. y = 110°(Corresponding angles)

x + y = 180° (Linear pair of angles)

or x + 110° = 180°

∴ x = 180° – 110° = 70°.13. An exterior angle of a triangle is equal

to the sum of two opposite interiorangles.∴ x = 30° + 50° = 80°.

14. Sum of the given angles= 40° + 60° + 80° = 180°

Yes, the triangle is possible.

15. Ratio = 3 km300 m

= 3 1000 m

300 m×

= 3000300

= 101

= 10 : 1.

16. ∵ Cost of 8 books = ` 240

∴ Cost of 1 book = `2408

= ` 30

∴ Cost of 15 books = ` 30 × 15= ` 450.

17. Number of present students = 32 – 8= 24

Required percentage

= Number of present students

Total number of students × 100

= 2432

× 100 = 34

× 100 = 75%.

18. Distance travelled in 1 hour

= Speed = Distance

Time =

89.12.2

= 89122

= 40.5 km.

Therefore, the bus travels 40.5 km in 1hour.

SECTION-C

19. We know that1 m = 100 cm or 100 cm = 1 m

Dividing both sides by 100, we have

1 cm = 1

m100

∴ 7 cm = 7 × 1

100 m = 0.07 m

Further,

100 cm = 1 m = 1

1000 km

(∵ 1000 m = 1 km)

∴ 1 cm =

11000100

km =

1100000

km

∴ 7 cm = 7

100000 km = 0.00007 km

Hence, 7 cm = 0.07 mand 7 cm = 0.00007 km.

20. Arranging the given data in ascendingorder, we get 12, 12, 13, 14, 14, 14, 14, 16, 19Mode:Mode of the data = The observation occuring mostly.

= 14.

186

Median:Number of observations = 9This is an odd number.

∴ Median = th9 1

2+

observation

= 5th observation= 14.

21.(i) 3p – 2 = 28or 3p = 28 + 2 = 30 (Transposing – 2 to RHS)

∴33p

= 303

(Dividing both sides by 3)∴ p = 10.

(ii) 0 = 20 + 5(m – 5)or 0 = 20 + 5m – 25 or 0 = 5m – 5or 5 = 5m (Transposing – 5 to LHS)

or 55

= m or 1 = m

i.e., m = 1.

22. The new solid (see figure) is a cuboid.

Length of the new solid,

l = 3 cm + 3 cm + 2 cm= 8 cm.

Breadth of the new solid, b = 2 cm

Height of the new solid, h = 2 cm.

23.(i) (122)3 ÷ 123 = (12)2 × 3 ÷ 123

= 126 ÷ 123

= (12)6 – 3 = 123.

(ii) (– 2)5 ÷ (– 2)3= (– 2)5 – 3

= (– 2)2 = [(– 1) × 2]2

= (– 1)2 × 22 = 1 × 22

= 22.24. Let the other number be x. Then

x + 1

14=

87

Subtracting 114

from both sides, we

get

x + 1

14 –

114

= 87

– 1

14

or x = 2 8 – 1

14×

= 16 – 1

14

∴ x = 1514

or 11

14.

Thus, the other number is 11

14.

25. P = ` 750, I = ` 225, R = 6%.

I = PRT100

∴ 225 = 750 6 T

100× ×

∴ T = 225 100750 6

××

= 225750

× 100

6

= 3

10 ×

503

= 5 years

Thus, the required time is 5 years.



26. (8 – 3y + 2y2) – (y2 + 6 – 4y)= 8 – 3y + 2y2 – y2 – 6 + 4y= (2y2 – y2) + (– 3y + 4y) + (8 – 6)= y2 + y + 2.

27. Let Salma’s present age be y years.∴ After 15 years, Salma’s age

= (y + 15) years.Also, after 15 years, Salma’s age

= 4 × Present age= 4y years.

∴ 4y = y + 15or 4y – y = 15 or 3y = 15

∴ y = 153

= 5.

Hence, Salma’s present age is 5 years.28. Let the two classes would get 5T and

7T toffees respectively.∴ 5T + 7T = 84or 12T = 84Divide both sides by 12, we get

T = 8412

= 7

∴ 5T = 5 × 7 = 35And 7T = 7 × 7 = 49Hence, the classes get 35 and 49 toffeesrespectively.

SECTION-D

29.(i)– 23

= – 2 43 4××

(∵ 123

= 4)

= – 812

.

(ii)– 47

= – 4 (–18)7 (–18)××

(∵ 72– 4

= – 18)

= 72

–126.

(iii)7

– 3=

7 (–1)– 3 (–1)××

= –73

.

(iv) Absolute form of – 824

= 8

24 =

13

.

(v)– 5– 20

= 520

(∵ ––

ab

= ab

)

And– 15

= – 1 45 4××

(∵ 205

= 4)

= – 420

.

30.(i) Let the man initially have ` x.

∴ Expenditure = ` 35

x

Now, according to question

x = 35

x + 1250 or x – 35

x = 1250

or 5 – 3

5x x

= 1250 or 25x

= 1250.


, we get

x = 1250 × 52

= 3125.

Therefore, the man initially has` 3125.

(ii)∵ Cost of 1 litre of petrol

= ` 4215

= ` 42 5 1

5× +

= ` 2115

∴ Cost of 1012

litres of petrol

= 1012

× `2115

= ` 21 2112 5

×

= ` 443110

= ` 443.10.

31.(i)23

× 1524

× 245

= 23

× 58

× 10 45+

= 23

× 58

× 145

= 28

× 55

× 143

188

= 14

× 1 × 143

= 7

2 × 3 =

76

= 116

.

(ii) 118

÷ 214

× 413

=1 1

1 28 4

÷ × 413

= 9 98 4

÷ × 133

= 9 48 9

× × 133

= 12

× 133

= 136

= 216

.

32.(i) In ∆ ABC,AB = AC

∴ x = 50°(Angles opposite to equal sides)

(ii) ∠2 + 120° = 180° (Linear pair of angles)∴ ∠2 = 180° – 120° = 60°.

Also, ∠1 = ∠2 = 60° (Angles opposite to equal sides)Further,∠1 + x = 120° (Exterior angle property) ∴ x = 120° – ∠1

= 120° – 60°(∵∠1 = 60° )

= 60°.(iii) In ∆ ABC,

AB = BC∴ x = ∠1Now, using angle sum property, we get

∠1 + x + 90° = 180°or x + x + 90° = 180° (∵ ∠1 = x )or 2x = 180° – 90° = 90°

∴ x = 902°

= 45°.

(iv) In ∆ ABC,

AC = BC∴ ∠1 = 20°Now, using angle sum property, weget

x + 20° + ∠1 = 180°or x + 20° + 20° = 180°

(∵∠1 = 20° )∴ x = 180° – 40° = 140°.

(v) Angles x and 70° are opposite toequal sides of the given triangle,∴ x = 70°.

33. (a) (i)The given net is of a right circularcylinder.

(ii) The given net is of a right circularcone.

(b) (i) In ∆ACO and ∆DBO,CO = DO (Given)

∠COA = ∠BOD (Vertically opposite angles)

AO = BO (Given)So, by SAS congruence criterion, wehave

∆ACO ≅ ∆BDO.(ii) In ∆ACO and ∆BDO,

CO = DO (Given)∠AOC = ∠BOD

(Vertically opposite angles)AO = BO (Given)

So, by SAS congruence criterion,∆ACO { ∆BDO.



34.(a) (i) This figure hasonly one line ofsymmetry passingthrough the pointof intersection ofPS and RQ.

(ii)

This figure has four lines ofsymmetry.

(b) A square has four lines of symmetry.

A hexagon has six lines of symmetry.


SECTION-A1. (B) (– 50) ÷ [(– 20) + (– 5)]

= (– 50) ÷ [– 20 – 5]= (– 50) ÷ (– 25)= 50 ÷ 25 = 2.

2. (A) Number of broken eggs

= 16

of 2 dozen

= 16

× 12 × 2 = 4.

3. (D) – 3– 7

= 37

, which is a positive

rational number.

4. (D) Perimeter = 13.34 cm.

5. (A) Rearranging the given data in thedescending order, we get 9, 8, 7, 6, 4, 3, 2n = Number of terms = 7, which isodd number.

∴ Median = th1

2n +

term

= th7 1

2+

term

= 4th term = 6.

6. (A)32

x = 15 or 32

x × 23

= 15 × 23

∴ x = 5 × 2 = 10.7. (D) The Pythagoras property holds for

a right-angled triangle.8. (C) Measures of two congruent angles

are equal.∴ Measure of other angle = 80°.

9. (B) Rectangle is a 2-D figure.10. (B) Zero is neither a positive nor a

negative number.

SECTION-B

11.(i) The given figure is a rhombus. So, ithas rotational symmetry of order 2.

(ii) The given figure is a regularpentagon. So, it has rotationalsymmetry of order 5.

12.(i) Sum of given angles = 90° + 55° + 35°= 180°.

Since the sum is 180°, Therefore, thetriangle is possible.

(ii) Sum of given angles = 50° + 50° + 61° = 161°.

Since the sum is not 180°, therefore,the triangle is not possible.

4.10 4.04+ 5.20 13.34

190

13. In the given figure, ∠AOC and ∠BOCform a linear pair.∴ ∠AOC + ∠BOC = 180°or 110° + ∠BOC = 180°∴ ∠BOC = 180° – 110° = 70°.

14. Decimal Form:

6.5% = 6.5100

= 65

1000 = 0.065.

Fractional Form:

6.5% = 6.5100

= 65

1000 =

13200

.

15. Let the required per cent be x%.Then,

x % of 42 = 7

or 100

x × 42 = 7

∴ x = 7 100

42×

= 100

6

= 1623

%.

16. 8y – 9 – 5y = 24or 3y – 9 = 24

(Transposing – 9 to RHS)or 3y = 24 + 9 = 33

or y = 333

= 11.

17. r = 56 cm.Area of the circle = pr2

= 227

× 56 × 56

= 176 × 56= 9856 cm2.

18. ∵ 1500 km covered in 30 litres

∴ 1 km covered in 30

1500litres

∴ 1800 km will cover in 30

1500 × 1800

litres

or 30 18

15×

litres or 36 litres.

Thus, 36 litres of petrol will be needed.

SECTION-C

19. Cost of 1 chair = Cost of 15 chairs

15

= 5532.30

15 =

55323150

= ` 368.82Cost of 21 chairs

= 21 × Cost of 1 chair= 21 × 368.82

= 21 36882

100×

= 774522

100= ` 7745.22.

Thus, the cost of 1 chair is ` 368.82 andthe cost of 21 chairs is ` 7445.22.

20. Outer radius = R = 20 mInner radius = r = 20 – 5 = 15 m

Area of the path= Area of the shaded region= Area of outer circle – Area

of inner circle.= pR2 – pr2 = p(R2 – r2)

= 227

(202 – 152)

= 227

× (400 – 225)

= 227

× 175 = 22 × 25

= 550.Thus, area of the path is 550 m2.



21. Area of a triangle =12

× Base × Altitude

∴ Altitude = 2 Area

Base×

= 2 90

30×

= 6 cm.22. Area of a square = (side)2 = (PQ)2

= PQ × PQ = 15 × 15= 225 cm2.

Perimeter of the square = 4 × PQ= 4 × 15= 60 cm.

23.

∆ABC is an equilateral triangle.∴ AB = BC = CA∆XYZ is also an equilateral triangle∴ XY = YZ = ZXBut AB = XY (Given)∴ BC = YZAnd CA = ZXSo, we conclude that ∆ABC and ∆XYZare congruent under SSS condition.

24.(i) A parallelogram has no line ofsymmetry.

(ii) An equilateral triangle has three linesof symmetry.

(iii) A semicircle has one line ofsymmetry.

25. Let the third angle be x:According to the angle sum propertyof a triangle, we have

50° + 50° + x = 180°or 100° + x = 180°Subtracting 100° from both the sides,we get

100° + x – 100° = 180° – 100°or x = 80°.Thus, the third angle is of measure 80°.

26. Line c is the transversal for the lines aand b.

Given angles of measures 34° and 136°are interior angles on the same side ofthe transversal c.Sum of these angles

= 34° + 136° = 170°Hence, the sum of interior angles onthe same side of the transversal is not180°, so the lines a and b are notparallel.

27. Let the angles be A, 2A and 6A.According to the angle sum propertyof a triangle,

A + 2A + 6A = 180°or 9A = 180°

∴ A = 180

9°

= 20°

∴ 2A = 2 × 20° = 40°and 6A = 6 × 20° = 120°.Thus, the measures of the angles ofthe given triangle are 20°, 40° and 120°.

28.(i) 3.7 × 4 = 3710

× 4 = 37 4

10×

= 14810

= 14.8.

(ii) 156.8 × 100 = 156810

× 100

= 1568 × 10 = 15680.

(iii) 2.835 ÷ 1000 = 28351000

× 1

1000

= 2835

1000000 = 0.002835.

192

SECTION-D

29. Let us re-arrange the given ages in theascending order, we get 35, 38, 43, 46, 46, 46, 47, 50, 52, 54(i) The oldest friend’s age = 54 years

The youngest friend’s age = 35 years(ii) Range = The oldest friend’s age –

The youngest friend’s age= 54 years – 35 years= 19 years.

(iii) Sum of all the ages= 35 + 38 + 43 + 46 + 46 + 46

+ 47 + 50 + 52 + 54= 457 years

∴ Mean age

=Sum of all the ages

Total number of the friends

= 45710

= 45.7 years.

(iv) Out of the given data, 46 years isoccured highest number of times

∴ Mode = 46 years.

30. (i) 911 ÷ 97 = 11

799

= 911 – 7 – =m

m nn

aa

a

∵

= 94.(ii) (a2 × x2)5 = (a2x2)5

= { }52( )ax [∵ pmqm = (pq)m]

= (ax)2 × 5

= (ax)10.(iii) (63)4 = 63 × 4 [∵ (ls)t = ls × t]

= 612.(iv) (– 2)4 × (– 2)– 4 = (– 2)4 – 4

[∵ am × an = am+n]= (– 2)0

= 1 [∵ a0 = 1 for a ! 0]

(v) ( )2507 = 7(50 × 2) [∵ (ls)t = ls × t]

= 7100.31.(i) 80,00,000 = 8.000000 × 1000000

= 8.0 × 106.(ii) 2,00,072 = 2 × 100000 + 0 × 10000

+ 0 × 1000 + 0 × 100 + 7 × 10 + 2 × 1= 2 × 105 + 0 + 0 + 0

+ 7 × 101 + 2 × 100

= 2 × 105 + 7 × 101 + 2 × 100.(iii) 3 × 3 × 3 × x × x × x × a × a × y × y ×y

= (3 × 3 × 3) × (x × x × x) × (a × a) × (y × y × y)

= 33 × x3 × a2 × y3

= (33 × x3 × y3) × a2

= (3 × x × y)3 × a2

[∵ a3 × b3 × c3 = (abc)3]= (3xy)3 × a2.

(iv)1

10000 81×

=1

(10 10 10 10) (3 3 3 3)× × × × × × ×

= 4 41

10 3×

= 41

(10 3)× [∵ am × bm = (a × b)m]

= 4

1(30)

= 30 – 4 –1

mm a

a = ∵

(v) (– 1)5 = (– 1) × (– 1) × (– 1) × (– 1) × (– 1)

= {(– 1) × (– 1)} × {(– 1) × (– 1)} × (– 1)

= 1 × 1 × (– 1) = 1 × (– 1)= – 1. [∵ a × (– 1) = – a]

32. (i) p + 7 = 18or p = 18 – 7 (Transposing 7 to RHS)∴ p = 11.



(ii) 5p – 12 = 28or 5p = 28 + 12 = 40


or55p

= 405

(Dividing both sides by 5)∴ p = 8.

(iii) 24 + 8(y – 8) = 0or 24 + 8y – 64 = 0or 8y – 40 = 0or 8y = 40∴ y = 5.

33. We know that x = x × 100%.

(i) 34

= 34

× 100% = 3 × 25% = 75%.

(ii) 2 : 8 = 28

= 14

= 14

× 100% = 25%.

(iii) 3.5 = 3.5 × 100% = 3510

× 100%

= 35 × 10% = 350%.34. (i) Let ABCD be a rhombus with

diagonals BD = 16 cm and AC = 30 cmLet AC and BD intersect each other atO.

We know that diagonals of a rhombusbisect each other at right angles.

∴ BO = DO = BD2

= 162

= 8 cm,

AO = CO = AC2

= 302

= 15 cm

And ∠COD = 90°In right triangle COD,

CD2 = CO2 + DO2

(Pythagoras property)= (15)2 + (8)2 = 225 + 64= 289.

∴ CD = 289 = 17 cm

Now perimeter of the rhombus= 4 × CD = 4 × 17= 68 cm.

Thus, the perimeter of the rhombus is68 cm.(ii) Radius r = 24.5 m

= 24510

m = 492

m

Circumference = 2pr

= 2 × 227

× 492

= 22 × 7 = 154 m.Distance covered in 4 complete turns

= 4 × Circumference= 4 × 154 = 616 m.

So, the distance covered by the boy is616 metres.


SECTION-A1. (A) Negative of – 7 = – (– 7) = 7

= Positive number.

2. (C) Let the fraction be xy

.

Then its reciprocal = yx

∴ xy

× yx

= xyxy

= 1.

3. (B) 1.444... is not a rational number.

4. (A) 345.50– 200.00 145.50

194

5. (C) Mode = Observation having highest frequency

= 5Range = 7 – 1 = 6.

6. (C) 5x + 3 = 18 or 5x = 18 – 3 = 15

∴ x = 155

= 3.

7. (C)∵ 92° + 44° + 44° = 180° and two ofangles are equal.So, this set of angles forms anisosceles triangle.

8. (C) 8 × 4x + 1 = 29

or 23 × (22)x +1 = 29

or 23 + 2 (x + 1) = 29

or 23 + 2x + 2 = 29

∴ 2x + 5 = 9

or x = 9 – 5

2=

42

= 2.

9. (C) This is the net for a hexagonalprism.

10. (A)– 57

= 28x

⇒ 7x = – 5 × 28

⇒ x = – 5 28

7×

= – 5 × 4 = – 20.

SECTION-B

11. 2pr = 8.8 or 2 × 227

× r = 8.8

∴ r = 8.8 72 22

××

= 8810

× 744

= 2 710×

= 75

= 1.4 m.

And 2r = 2 × 1.4 = 2.8 m.Thus, diameter = 2.8 mand radius = 1.4 m.

12. An exterior angle of a triangle is equalto the sum of two opposite interiorangles∴ x + 60° = 130°or x = 130° – 60°

(Transposing 60° to RHS)= 70°.

13. P = ` 184, R = 5%, T = 2 years

I = PRT100

= 184 5 2

100× ×

= 184 1010 10

××

= 18410

= 18.4.

Thus, the interest be ` 18.4.14. Let R% of 1 km be 75 m.

∴ R100

× 1 km = 75 m

orR

100 × 1000 m = 75 m

[1 km = 1000 m]

∴ R = 75 100

1000×

= 7510

= 7.5.

Thus, the required percentage is 7.5%.

15. 0.4 : 0.6 = 0.40.6

= 0.4 100.6 10

××

= 46

= 23

= 2 : 3.16. x + 5x + 7 = 25

or 6x + 7 = 25or 6x = 25 – 7 = 18

or66x

= 186


∴ x = 3.



17. (i) Degree of x2 + xyz + y3

= Degree of xyz or degree of y3

= 3.

(ii) Degree of m2n2 + mn2 + 2

= Degree of m2n2.

= 2 + 2 = 4.

18. First 5 natural numbers are:

1, 2, 3, 4, 5

Sum of these numbers

= 1 + 2 + 3 + 4 + 5 = 15

∴ Mean= 155

= 3.

SECTION-C

19. Rearranging the given salaries in theascending order, we have` 38, ` 40, ` 42, ` 45, ` 50, ` 50, ` 60,` 71, ` 82, ` 84, ` 90.Number of the salaries = 11This is an odd number.

∴ Median = th11 1

2+

term

= th12

2 term

= 6th term = ` 50.20. Area = 250 m2, Base (b) = 50 m,

Altitude (h) = ?

Area = 12

bh

∴ h = 2 Area

b×

= 2 250

50×

= 2 × 5 = 10 m.Thus, the altitude is 10 metres.

21. Perimeter of a square = 4 × Side∴ 48 = 4 × Side

∴ 484

= Side

or 12 = SideArea of the garden = (side)2

= 12 × 12= 144 m2.

22. AB y CD and PQ is transversal∴ x = 50° (Corresponding angles)

AB y EF and PQ is transversal∴ y = x

(Alternate interior angles)= 50° (∵ x = 50°)

Thus, x = y = 50°.23. SP for each horse = ` 324.

CP of the horse which provides gain

= SP 100

(100 20)×+

= 324 100

120×

= 324012

= ` 270.CP for the horse which provides loss

= SP 100

(100 – 20)×

= 324 100

80×

= 3240

8= ` 405.

∴ Total CP = 270 + 405= ` 675

Total SP = 2 × 324 = ` 648.Since, total CP is greater than total SP.So, there is a loss in the wholetransation.∴ Loss = CP – SP = 675 – 648

= ` 27.

196

24. P = ` 300, A = 2 × 300 = ` 600, R = 4%∴ I = A – P = 600 – 300

= ` 300

I = PRT100

or I = I 100

PR×

∴ T = 300 100300 4

××

= 100

4= 25 years.

Thus, the money will double itself in25 years.

25. Ratio of 7 to 11

= 711

= 7 311 3××

= 2133

= 21 : 33

Ratio of 21 to 33 = 2133

= 21 : 33

Since 7 : 11 = 21 : 33Therefore, 7, 11, 21, 33 are inproportion.

26. ∵ Weight of 405 book = 90 kg

∴ Weight of 1 book = 90405

kg

= 29

kg.

∴ (i) weight of 540 books

= 29

× 540 kg

= 2 × 60 kg = 120 kg.(ii) Required number of books

= 50 kg

Weight of 1 book

= 50 kg2

kg9

= 5029

= 50 × 92

= 25 × 9

= 225.

27. Total CP= 520000 + 80000 = ` 600000SP = ` 640000

∵ SP > CP ∴ There is a profit.Profit = 640000 – 600000 = ` 40000

Profit per cent = Profit

Total CP × 100

= 40000

600000 × 100

= 40

%6

= 623

%

Thus, the profit per cent is 623

%.

28. Substituting a = – 2 and b = 1 in(i) a2 – b2, we get

a2 – b2 = (– 2)2 – (1)2 = 4 – 1 = 3Thus, a2 – b2 = 3.

(ii) a + b, we geta + b = (– 2) + 1 = – 2 + 1 = – 1Thus, a + b = – 1.

SECTION-D

29. Let the other number be p. Then

p × – 26

8 =

158


– 26, we

get

p × – 26

8 ×

8– 26

= 158

× 8

– 26

or p = 15

– 26=

–1526

Thus, the other number is –1526

.

30. Add x2 – 3xy + y2 and x2 + 5xy – y2 as

2 2

2 2

2

– 3 ++ + 5 –

2 + 2

x xy yx xy yx xy



Subtract x2 – 4xy + 4y2 from 2x2 + 2xyas

31. (i) ABD is a triangle.∴ Sum of its interior angles = 180°i.e., a + 100° + 40° = 180°or a + 140° = 180°∴ a = 180° – 140°

(Transposing 140° to RHS)= 40°

Similarly, for ∆BCD,b + 80° + 30° = 180°

or b + 110° = 180°∴ b = 180° – 110° = 70°

(Transposing 110° to RHS)Thus, a = 40° and b = 70°.(ii) In ∆ABC and ∆FED,

∠B = ∠E (Each right angle)AC = DF (Hypotenuse)BC = DE (Sides)

So, by RHS congruence criterion,∆ABC { ∆FED.

32. (i) Let the whole quantity be x. Then

5% of x = 800 or 5

100 × x = 800

∴ x = 800 100

5×

= 160 × 100 = 16000Thus, the whole quantity is 16000.(ii) Total number of students = 50Number of girls = 40% of 50

= 40

100 × 50

= 4 × 5 = 20.Number of boys

= Total number of students– Number of girls

= 50 – 20 = 30.Thus, the number of boys is 30.(iii) Total number of students = 40Number of students who like playing

football = 25% of 40 = 25

100 × 40

= 1000100

= 10

∴ Number of students who do notlike playing football = 40 – 10 = 30.Thus, 30 students do not like playingfootball.

33. (i) Subtract 1115

from –1320

.

Let us first find LCM of 15 and 20.

∴ LCM (15, 20)= 2 × 2 × 3 × 5 = 60

∴ 1115

= 11 415 4

××

= 4460

60

415

∵ =

and–1320

= –13 320 3

××

60

320

= ∵

= – 3960

Now,–1320

– 1115

= – 3960

– 4460

= –39 4460 60

+

= –39 44

60+

= – 8360

.

2

2 2

2 2

2 + 2+ + 4 + 4

+ 6 – 4

x xyx xy yx xy y

2 15, 202 15, 103 15, 55 5, 5

1, 1

198

(ii) 23

14=

2 14 314

× + =

28 314+

= 3114

– 3– 7

= 37

Add 3114

and 37

.

Let us find LCM of 7 and 14.

∴ LCM (7, 14)

= 2 × 7 = 14

∴ 37

= 3 27 2××

= 6

14

14

= 27

∵

Now, 23

14 +

– 3–7

= 3114

+ 614

= 31+ 6

14 =

3714

= 29

14.

34. (i) RECTANGLE:Area = 40 cm2, base = 5 cmWe know that:Area of a rectangle = Base × Height∴ 40 = 5 × Height

∴ Height = 405

= 8 cm.

TRIANGLE:Base = b = 5 cmHeight = h = height of the rectangle

= 8 cm.

∴Area of the triangle = 12

bh

= 12

× 5 × 8

= 5 × 4= 20 cm2.

Thus, area of the triangle is 20 cm2.

(ii) Length of rectangle (l) = 24 mBreadth of rectangle (b) = 5 m∴ Area of rectangle = l × b

= 24 × 5 m2

Side of square = 2 m∴ Area of square = (side)2

= 2 × 2 = 4 m2

Now, number of squares

= Area of the rectangle

Area of a square

= 24 × 5

4 = 6 × 5 = 30.

Thus, 30 squares can be cut from theflower bed.

Practice Paper – 5

SECTION-A

1. (C) 7 is a positive integer

7 × (– 1) = – 7 = Negative integer

2. (B) 338

= 3 8 3

8× +

= 278

∴ Multiplicative inverse = 1

278

= 8

27.

3. (A) ∵ – 1 > – 3 > – 9

∴ – 117

> – 317

> – 917

∴ Required number =– 917

– –117

= – 917

+ 1

17

= – 817

.

4. (B) 0.07 × 7.08= 0.07 100

100×

×7.08 100

100×

2 7, 147 7, 7

1, 1



= 7

100 ×

708100

= 4956

10000= 0.4956.

5. (C) Mean

= 1 2 3 4 5 6 7 8 9 10

10+ + + + + + + + +

= 5510

= 5.5.

6. (C) Let the unknown number be x.Then,

5 + 32

of x = 20 or 5 + 32

x = 20

or 10 + 3x = 40 or x = 303

∴ x = 10.7. (A) An exterior angle = Sum of two

opposite interior angles∴ ∠1 = ∠2 + ∠3.

8. (A)

∴ 67375 = 53 × 72 × 11

= 53 × (– 7)2 × 11.9. (D) A sphere has no vertex.

10. (A) The integer 7 can be re-written as

71

which is a rational number.

SECTION-B

11. (i) Product of 6 negative integers = Positive integer.

Positive integer × Positive integer = Positive integer

Thus., the product of 6 negativeintegers and 1 positive integer willhave positive sign.(ii) Product of 19 negative integers = Negative integerProduct of 3 positive integers

= Positive integerProduct of 1 negative integer and 1positive integer = Negative integerThus, the product of 19 negativeintegers and 3 positive integers willhave negative sign.

12. l = 8 cm,

b = 312

cm = 6 1

2+

cm = 72

cm

Area of the rectangle = l × b = 8 × 72

= 562

= 28 cm2.

13.(i)∵ 1 km = 1000 m∴ 8 km = 8 × 1000 m = 8000 m∴ 8 km 35 m = 8000 m + 35 m

= 8035 m.(ii)∵ 1000 mm = 1 m

∴ 1 mm = 1

1000 m

∴ 4509 mm = 1

1000 × 4509

= 4.509 m.

14.(i)– 80100

= 80

– 100

= 4 20

– 5 20××

= 4

– 5.

(ii) – 8211

= – 2

811

= –8 11 2

11× +

= – 88 2

11+

= –9011

= – 9011

.

5 673755 134755 26957 5397 7711 11

1

200

15.(i) 12

– 14

= 1 2 – 1 1

4× ×

= 2 – 1

4 =

14

Reciprocal of 1 1

–2 4

= Reciprocal of 14

= 4.

(ii)78

× – 320

= 7 (– 3)8 20××

= – 21160

Reciprocal of – 37

8 20 ×

= Reciprocal of – 21160

= 160– 21

= –160

21.

16.(i) (– 2)4 × (– 2)13 = (– 2)4 + 13

(∵ am × an = am+n)= (– 2)17.

(ii)

32 4– 2 2–

3 5

×

=2 3– 2

3

× × 4 3– 2

5

×

= 6– 2

3 ×

12– 25

= ( )6

6– 23

× ( )12

12– 25

= 18

6 12

(– 2)3 5×

= 218 × 3– 6 × 5 – 12.17. Substituting a = – 4 in 7a2 + 7a – 5, we

get7a2 + 7a – 5 = 7(– 4)2 + 7(– 4) – 5

= 7 × 16 – 28 – 5= 112 – 33 = 79.

18. 4t + 5 = t + 15Transposing 5 to RHS and t to LHS,we get

4t – t = 15 – 5or 3t = 10Dividing both sides by 3, we get

33t

= 103

or t = 103

= 1

33

.

SECTION-C

19. Let the numbers be 7x and 13x.Their sum = 7x + 13x = 20x

According to the given conditions, wehave

20x = 980

or x = 98020

= 49

(Dividing both sides by 20)∴ 7x = 7 × 49 = 343and 13x = 13 × 49 = 637Hence, the required numbers are 343and 637.

20. Capacity of 1 box

= Capacity of 45 boxes

45

= 283545

= 3155

= 63 laddoos∴ Required number of boxes to fill6615 laddoos

= 6615

Capacity of 1 box

= 661563

= 7357

= 105Thus, 105 boxes will be required.

21.(i) Number of days in the month ofApril = 30



∴ 40% of the days in the month ofApril = 40% of 30

= 40

100 × 30 = 4 × 3

= 12 days.

(ii) 30% of 800 = 30

100 × 800 = 30 × 8

= 240.Decreasing 240 from 800, we get

800 – 240 = 560.22. CP = ` 2005

Profit = 10% of CP

= 10100

× 2005 = ` 200.5

Now, SP = CP + Profit= ` 2005 + ` 200.5 = ` 2205.50.

ORLet SP = ` x

CP = ` 2005, Profit% = 10%.Profit = SP – CP = x – 2005

We have, profit per cent

= Profit

CP × 100

∴ 10% = 2005

2005x ×

× 100

∴ 10 2005100×

= x – 2005

or 200.5 = x – 2005or x = 2005 + 200.5

= 2205.50Therefore, the selling price is ` 2205.50.

23. P = ` 250,A = 2 × P = 2 × 250 = ` 500,

∴ I = A – P = 500 – 250 = ` 250R = 8%

We know that

I = PRT100

∴ T = I 100P R××

= 250 100250 8

××

= 100

8= 12.5 years.

Thus, the required number of years is12.5.

24.(i) Let the given angle be 45° + xHere, x is greater than zero and lessthan 45°.Then its complement

= 90° – (45° + x)= 90° – 45° – x= 45° – x

which is less than 45°.Thus, the required complement angleis less than 45°.

(ii) The angles shown in the figure forma linear pair of angles.∴ 2x + 3x = 180°or 5x = 180°

or 55x

= 1805°

(Dividing both sides by 5)∴ x = 36°.

25. Angle sum property of a triangle: Thetotal measures of angles of a triangleis 180°

(i) Let the third angle be x. Thenx + 40° + 80° = 180°

or x + 120° = 180°∴ x = 180° – 120° = 60°Thus, the measure of third angle is60°.

(ii) Let the third angle be y. Theny + 25° + 114° = 180°

or y + 139° = 180°∴ y = 180° – 139° = 41°Thus, the measure of the third angleis 41°.

202

26.(i) The order of rotational symmetry ofa square is 4.

(ii)

There is 1 line of symmetry for letter B.(iii)

27.

In ∆ ABC and ∆ DEF,

AB = DE (Given)

∠B = ∠E (Given)

BC = EF (Given)

So, by SAS congruence condition, wehave

∆ ABC { ∆ DEF.

28. Breadth b = 90 cm,

Perimeter = 400 cm, Length l = ?

Perimeter of a rectangle = 2(l + b)

∴ 400 = 2(l + 90)

or 400 = 2l + 180

∴ 400 – 180 = 2l

or2202

= l

or l = 110

Thus, the length of the rectangle is110 cm.

SECTION-D

29. Length l = 100 m, Breadth b = 45 m(i) The playground is in the form of a

rectangle.∴Area of the playground

= l × b= 100 × 45= 4500 m2

Cost of levelling= Area × Cost per m2

= 4500 × 5.50= 45 × 550 = 24750.

Thus, the cost of levelling theplayground is ` 24750.

(ii) Perimeter of the playground= 2(l + b)= 2(100 + 45)= 290 m.

Distance covered by the boy= 4 × Perimeter= 4 × 290 = 1160 m.

Speed = 5.8 m/minute

Time taken = Distance

Speed

= 11605.8

= 11600

58= 200 minutes.

or 3 hr 20 minutes.30. Let a be the side of the square and b be

the base of the parallelogram.

SquarePerimeter of the square = 4 × a



But the given perimeter of the square= 250 m

∴ 4a = 250

∴ a = 1252

m

Height of the parallelogram h = 50 m

Parallelogram

Area of the parallelogram= Area of the square

(Given)or b × h = a × a

∴ b × 50 = 1252

× 125

2Dividing both sides by 50, we get

or50

50b ×

= 125 12550 2 2

×× ×

∴ b = 6258

= 78.125 m.

Thus, the measure of the correspondingbase of the parallelogram is 78.125 m.

31.(i) When you change a closed figure toanother closed figure, the perimeterremain unchanged.

Area of the square = 121 cm2

∴ Side × Side = 11 × 11∴ Side = 11 cm.Circumference of the circle

= Perimeter of the squareor 2pr = 4 × Side

or 2 × 227

× r = 4 × 11

∴ r = 4 11 7

22 2× ×

× = 7 cm.

Area of the circle

= pr2 = 227

× 7 × 7

= 22 × 7 = 154 cm2

Thus, area of the circle is 154 cm2.

(ii) Area of the circle = 15400 m2

or pr2 = 15400

∴ r2 = 15400p

= 15400

227

= 15400 7

22×

= 700 × 7

or r × r = (7 × 10) × (7 × 10)

∴ r = 7 × 10 = 70 m

∴ Diameter = 2 × r = 2 × 70

= 140 m.

Thus, the diameter of the circle is140 metres.

32. To draw a double bar graph, you haveto go to the steps:

Step I. Draw a pair of perpendicularlines OX and OY on a graph paper.

Step II. Along the horizontal axis (OX),mark the days of the week, namelyMon, Tue, Wed, Thur and Fri. Alongthe vertical axis (OY), mark the numberof absentees.

Step III. Choose a suitable scale todetermine the height of bars. Here take

1 absentee = 1 big division on OY.

Step IV. First draw the bars for week 1and then for week 2 by taking equalwidth of the bars and equal gapbetween any two consecutive bar pairs.

Step V. Shade the bars of the weekswith different types. Show their

204

shadings on the top right corner of thegraph paper.

33.(i) 120° and ∠1 form a linear pair.

∴ ∠1 + 120° = 180°

∴ ∠1 = 180° – 120° = 60°

Now, x = ∠1 (Angles opposite to equal sides)

= 60°.(ii) 50° and ∠1 are vertically opposite

angles

∴ ∠1 = 50°x and ∠1 are opposite to equal sides,∴ x = ∠1 = 50°.

(iii) x = y (Angles opposite to equal sides)

x + y = 100° (Exterior angle property)

or x + x = 100° (∵ x = y)

or 2x = 100°

∴ x = 1002°

= 50°.

(iv) 60° + x = 180° (Linear pair of angles)∴ x = 180° – 60°

(Transposing 60° to RHS)= 120°.

(v) Using exterior angle property for atriangle, we havex + 30° = 110°

∴ x = 110° – 30°(Transposing 30° to RHS)

or x = 80°.34.(i) Let Raghu’s age be x years

Two times of x = 2 × x = 2x5 more than 2x = 2x + 5

Consequently, we get2x + 5 = 41 or 2x = 41 – 5

or 2x = 36 or x = 362

or x = 18Therefore, Raghu’s age is 18 years.

(ii) In 1 hour, Bulbul reads 13

part

In 2 16

hours she will read 12

6 × 1

3

part 136

× 13

part or 1318

part.

Thus, Bulbul will read 1318

part in

216

hours.

❏❏


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